Trigonometric Equation Exercise: 8.2 Class 11 Basic Mathematics Solution [NEB UPDATED]

 

Exercise 8.2

Solve the following equations from exercise 1 to 10.

1.

a. 4cos²x = 1

Solution:

Here, 4cos²x = 1

Or, cos²x = $\frac{1}{4}$

Or, cos²x = ${\left( { \pm \frac{1}{2}} \right)^2}$

Or, cos²x = cos²$\frac{{\rm{\pi }}}{3}$.

So, x = nπ $ \pm $$\frac{{\rm{\pi }}}{3}$, n ԑ Z    [cos²x = cos²α àx = nπ $ \pm $ α]

 

b. cos 2x – sinx = 0

Solution:

Here, cos 2x – sinx = 0

Or, cos 2x = cos $\left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)$

So, 2x = 2nπ $ \pm $$\left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)$

Or, 2x = 2nπ + $\frac{{\rm{\pi }}}{2}$– x           and    2x = 2nπ – $\frac{{\rm{\pi }}}{2}$ + x

Or, 3x = (4n + 1) $\frac{{\rm{\pi }}}{2}$             and   x = (4n – 1) $\frac{{\rm{\pi }}}{2}$.

So, x = (4n + 1)$\frac{{\rm{\pi }}}{6}$, (4n – 1)$\frac{{\rm{\pi }}}{2}$, n ԑ Z.

 

c. 2 sin2x = sinx

Solution:

Here, 2 sin2x = sinx

Or, 4sinx cosx – sinx = 0

Or, sinx (4cosx – 1) = 0

Either, sinx = 0

So, x = nπ

Or, 4cosx – 1 = 0

Or, cosx = $\frac{1}{4}$

Or, cosx = cosα [cosx = $\frac{1}{4}$]

So, x = 2nπ $ \pm $ α.

So, x = 2nπ $ \pm $ cos^–1$\frac{1}{4}$ n ԑ Z.

 

d. sin2x + sinx = 0

Solution:

Here, sin2x + sinx = 0

Or, 2 sinx.cosx + sinx = 0

Or, sinx(2cosx + 1) = 0

Either, sinx = 0

So, x = nπ.

Or, 2cosx + 1 = 0

Or, cosx = $ - \frac{1}{2}$ = cos$\frac{{2{\rm{\pi }}}}{3}$

So, x = 2nπ $ \pm $$\frac{{2{\rm{\pi }}}}{3}$ = (6n$ \pm $ 2) $\frac{{\rm{\pi }}}{3}$

Hence, x = nπ, (6n$ \pm $ 2) $\frac{{\rm{\pi }}}{3}$, n ԑ Z.

 

2.

a. sin²x – cosx = 1

Solution:

Here, sin²x – cosx = 1

Or, –cosx(cosx + 1) = 0.

So, cosx(cosx + 1) = 0

Either, cos x = 0

Or, cosx = cos(2n + 1) $\frac{{\rm{\pi }}}{2}$

So, x = (2n + 1) $\frac{{\rm{\pi }}}{2}$.

Or, cosx + 1 = 0

Or, cosx = –1.

So,cosx = cos(2n + 1)π

So, x = (2n + 1)π.

Hence, x = (2n + 1) $\frac{{\rm{\pi }}}{2}$, (2n + 1)π, n ԑ Z.

 

b. cos²x – sinx + 5 = 0

Solution:

Here, cos²x – sinx + 5 = 0

Or, 1 – sin²x – sinx + 5 = 0

Or, sin²x + sinx – 6 = 0

Or, sin²x + 3sinx – 2sinx – 6= 0

Or, sinx(sinx + 3) – 2(sinx + 3) = 0

Or, (sinx + 3)(sinx – 2) = 0

Either, sinx – 2 = 0

SO, sinx = 2 > 1.

Or, sinx + 3 = 0

So, sinx = –3 <–1.

Since, –1 ≤ sinx ≤ 1, the given equation does not have a solution.

 

c. 4cosx + secx – 4 = 0

Solution:

Here, 4cosx + secx – 4 = 0

Or, 4cos²x + 1 – 4cosx = 0

Or, 4cos²x – 4cosx + 1 = 0

Or, (2cosx – 1)² = 0.

Or, 2cosx – 1 = 0

Or, cos x = $\frac{1}{2}$.

Or, cosx = cos $\frac{{\rm{\pi }}}{3}$

So, x = $\left( {2{\rm{n\pi }} \pm \frac{{\rm{\pi }}}{3}} \right),$n ԑ Z.

 

d. 2sinx + cotx – cosec x = 0

Here, 2sinx + cotx – cosec x = 0

Or, 2sinx + $\frac{{{\rm{cosx}}}}{{{\rm{sinx}}}}$ – $\frac{1}{{{\rm{sinx}}}}$ = 0

Or, 2sin²x + cosx – 1 = 0.

Or, 2(1 – cos²x) + cosx –1 = 0

Or, –2cos²x + cosx + 1 = 0

So, 2cos²x – cosx – 1 = 0

Or, 2cos²x – 2cosx + cosx – 1 = 0

Or, 2cosx (cosx – 1) + 1(cosx – 1) = 0

Or, (cosx – 1)(2cosx + 1) = 0

Either, 2cosx + 1 = 0

So, cos x = $ - \frac{1}{2}$

Or, cos x = cos$\frac{{2{\rm{\pi }}}}{3}$

So, x = 2nπ $ \pm \frac{{2{\rm{\pi }}}}{3}$ = (6n $ \pm $ 2) $\frac{{\rm{\pi }}}{3}$.

Or, cosx – 1 = 0

Or, cosx = 1

Or, cosx = cos 0

So, x = 2nπ $ \pm $ 0 = 2nπ

Hence, x = (6n $ \pm $ 2) $\frac{{\rm{\pi }}}{3}$, 2nπ, n ԑ Z.

 

3.

a. 2cos²x + 4sin²x = 3

Solution:

Here, 2cos²x + 4sin²x = 3

Or, 2cos²x + 4(1 – cos²x) – 3 =  0

Or, 2cos²x + 4 – 4 cos²x – 3 = 0

Or, –2cos²x + 1 = 0

Or, cos² x = $\frac{1}{2}$.

Or, cos²x = ${\left( { \pm \frac{1}{{\sqrt 2 }}} \right)^2}$ = cos²$\frac{{\rm{\pi }}}{4}$

So, x = nπ $ \pm $$\frac{{\rm{\pi }}}{4}$, n ԑ Z.

 

b. 7sin²x + 3cos²x = 4

Solution:

Here, 7sin²x + 3cos²x = 4

Or, 7(1 – cos²x) + 3cos²x – 4 = 0

Or, 7 – 7 cos²x + 3cos²x – 4 = 0

Or, –4cos²x + 3 = 0

Or, cos²x = $\frac{3}{4}$ = $\left( { \pm \frac{{\sqrt 3 }}{2}} \right)$ = cos²$\frac{{\rm{\pi }}}{6}$.

So.  x = nπ, $ \pm \frac{{\rm{\pi }}}{6}$, n ԑ Z.

 

c. tanx + cotx - 2cosec x =0

Solution:

Here, tanx + cotx = 2cosec x

Or, $\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$ + $\frac{{{\rm{cosx}}}}{{{\rm{sinx}}}}$ = $\frac{2}{{{\rm{sinx}}}}$

Or, $\frac{{{{\sin }^2}{\rm{x}} + {{\cos }^2}{\rm{x}}}}{{{\rm{sinx}}.{\rm{cosx}}}}$ = $\frac{2}{{{\rm{sinx}}}}$

Or, $\frac{1}{{{\rm{sinx}}.{\rm{cosx}}}} = \frac{2}{{{\rm{sinx}}}}$

Or, 2sinx.cosx = sinx.

Or, sinx(2cosx – 1) = 9

Either, sinx = 0 = sin 0

So, x = nπ

Or, 2 cosx – 1 = 0

Or, cosx = $\frac{1}{2}$ = cos$\frac{{\rm{\pi }}}{3}$

SO, x = 2nπ $ \pm $$\frac{{\rm{\pi }}}{3}$ = (6n $ \pm $ 1) $\frac{{\rm{\pi }}}{3}$.

Hence, x = nπ, (6n$ \pm $1) $\frac{{\rm{\pi }}}{3}$, n ԑ Z.

 

d. tan²x = secx + 1

Solution:

Here, tan²x = secx + 1.

Or, sec²x – 1 = sec x + 1

Or, sec²x – secx – 2 = 0

Or, sec²x – 2secx + secx – 2 = 0

Or, secx(secx – 2) + 1(secx – 2) = 0

Or, (secx – 2)(secx + 1) = 0

Either, secx – 2 = 0

So, sec x = 2

Or, cosx = $\frac{1}{2}$ = cos$\frac{{\rm{\pi }}}{3}$

So, x = 2nπ $ \pm \frac{{\rm{\pi }}}{3}$

Or, sec + 1 = 0

Or, sec = –1

Or, cosx = –1

So, x = (2n+1)π

Hence, x = (2n + 1)π, (6n$ \pm $1) $\frac{{\rm{\pi }}}{3}$.

 

4.

a. sinx + $\sqrt 3 $cosx = $\sqrt 2 $

Solution:

Here, sinx + $\sqrt 3 $cosx = $\sqrt 2 $

Or, Dividing by $\sqrt {{1^2} + {{\sqrt 3 }^2}} $ = 2, we have,

Or, $\frac{1}{2}$sinx + $\frac{{\sqrt 3 }}{2}$cosx = $\frac{{\sqrt 2 }}{2}$

Or, sin $\frac{{\rm{\pi }}}{6}$ sinx + cos $\frac{{\rm{\pi }}}{6}$ cosx = $\frac{1}{{\sqrt 2 }}$

Or, cos $\left( {{\rm{x}} - \frac{{\rm{\pi }}}{6}} \right)$ = cos $\frac{{\rm{\pi }}}{4}$

So, x – $\frac{{\rm{\pi }}}{6}$ = 2nπ $ \pm \frac{{\rm{\pi }}}{4}$.

Hence, x = 2nπ + $\frac{{\rm{\pi }}}{6}$$ \pm $$\frac{{\rm{\pi }}}{4}$, n ԑ Z.

 

b. $\sqrt 3 $sin – cosx = $\sqrt 2 $

Solution:

Or, $\sqrt 3 $sin – cosx = $\sqrt 2 $

Or, $\frac{{\sqrt 3 }}{2}$sinx – $\frac{1}{2}$ cosx = $\frac{1}{{\sqrt 2 }}$

Or, sin $\frac{{\rm{\pi }}}{3}$ sinx – cos$\frac{{\rm{\pi }}}{3}$ cosx = $\frac{1}{{\sqrt 2 }}$

Or, –cos $\left( {{\rm{x}} + \frac{{\rm{\pi }}}{3}} \right)$ = $\frac{1}{{\sqrt 2 }}$

Or, cos$\left( {{\rm{x}} + \frac{{\rm{\pi }}}{3}} \right)$ = $ - \frac{1}{{\sqrt 2 }}$ = cos $\frac{{3{\rm{\pi }}}}{4}$

So, x + $\frac{{\rm{\pi }}}{3}$ = 2nπ $ \pm \frac{{3{\rm{\pi }}}}{4}$.

So, x = 2nπ – $\frac{{\rm{\pi }}}{3}$$ \pm \frac{{3{\rm{\pi }}}}{4}$, n ԑ Z.

Since, 0 ≤ x ≤ π, So,

For n = 0,

X = $ - \frac{{\rm{\pi }}}{3} \pm \frac{{3{\rm{\pi }}}}{4}$ = $ - \frac{{\rm{\pi }}}{3} + \frac{{3{\rm{\pi }}}}{4}, - \frac{{\rm{\pi }}}{3} - \frac{{3{\rm{\pi }}}}{4}$ = $\frac{{5{\rm{\pi }}}}{{12}}$

For, n = 1.

X = $2{\rm{\pi }} - \frac{{\rm{\pi }}}{3} \pm \frac{{3{\rm{\pi }}}}{4}$ = $2{\rm{\pi }} - \frac{{\rm{\pi }}}{3} + \frac{{3{\rm{\pi }}}}{4},2{\rm{\pi }} - \frac{{\rm{\pi }}}{3} - \frac{{3{\rm{\pi }}}}{4}$

= $\frac{{24{\rm{\pi }} - 4{\rm{\pi }} + 9{\rm{\pi }}}}{{12}}$, $\frac{{24{\rm{\pi }} - 4{\rm{\pi }} - 9{\rm{\pi }}}}{{12}}$

= $\frac{{29{\rm{\pi }}}}{{12}}$, $\frac{{11{\rm{\pi }}}}{{12}}$

Since, x = $\frac{{29{\rm{\pi }}}}{{12}}$ does not lie in [0,π]

So, x = $\frac{{5{\rm{\pi }}}}{{12}},\frac{{11{\rm{\pi }}}}{{12}}.$

 

c. cosx + $\sqrt 3 $sinx = $\sqrt 2 $

Solution:

Here, cosx + $\sqrt 3 $sinx = $\sqrt 2 $

Or. $\frac{1}{2}$ cosx + $\frac{{\sqrt 3 }}{2}$sinx = $\frac{{\sqrt 2 }}{2}$     [Dividing by $\sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}} $= 2 ]

Or, cos $\frac{{\rm{\pi }}}{3}$ cosx + sin $\frac{{\rm{\pi }}}{3}$ sinx = $\frac{1}{{\sqrt 2 }}$

Or, cos $\left( {{\rm{x}} - \frac{{\rm{\pi }}}{3}} \right)$ = cos $\frac{{\rm{\pi }}}{4}$

Or, x – $\frac{{\rm{\pi }}}{3}$ = 2nπ $ \pm $$\frac{{\rm{\pi }}}{4}$

Hence, x = 2nπ + $\frac{{\rm{\pi }}}{3}$$ \pm \frac{{\rm{\pi }}}{4}$, n ԑ Z.

 

d. sinx + cosx = $\sqrt 2 $

Solution:

Here, sinx + cosx = $\sqrt 2 $

Or, $\frac{1}{{\sqrt 2 }}$sinx + $\frac{1}{{\sqrt 2 }}$cosx = $\frac{{\sqrt 2 }}{{\sqrt 2 }}$      [Dividing $\sqrt {{1^2} + {1^2}} $ = $\sqrt 2 $]

Or, cosx cos $\frac{{\rm{\pi }}}{4}$ + sinx sin$\frac{{\rm{\pi }}}{4}$ = 1

Or, cos$\left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)$ = cos0

So, x – $\frac{{\rm{\pi }}}{4}$ = 2nπ $ \pm $ 0.

So, x = 2nπ + $\frac{{\rm{\pi }}}{4}$, n ԑ Z  

Since, –2π ≤ x ≤ 2π, so

For n = 0, x = $\frac{{\rm{\pi }}}{4}$

For, n = –1, x = –2π + $\frac{{\rm{\pi }}}{4}$ = $ - \frac{{7{\rm{\pi }}}}{4}$

Hence, x = $\frac{{\rm{\pi }}}{4}$, $ - \frac{{7{\rm{\pi }}}}{4}$.

 

5.

a. sin3x + sinx = 0

Solution:

Here, sin3x + sinx = 0.

Or, 2 sin$\frac{{3{\rm{x}} + {\rm{x}}}}{2}$. cos$\frac{{3{\rm{x}} - {\rm{x}}}}{2}$ = 0.

Or, sin2x.cosx = 0

Either, sin2x = 0

So, 2x = nπ

So, x $ = \frac{{{\rm{n\pi }}}}{2}$

Or. Cosx = 0 = cos $\frac{{\rm{\pi }}}{2}$.

So, x = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$ = (4n $ \pm $ 1) = $\frac{{\rm{\pi }}}{2}$.

Hence, x = $\frac{{{\rm{n\pi }}}}{2}$, (4n$ \pm $1) $\frac{{\rm{\pi }}}{2}$.

 

b. sin3x + sinx = sin2x

Solution:

Here, sin3x + sinx = sin2x

Or, 2 sin$\frac{{3{\rm{x}} + {\rm{x}}}}{2}$. cos$\frac{{2{\rm{x}} - {\rm{x}}}}{2}$– sin2x = 0

Or, 2sin2x. cosx – sin2x = 0

Or, sin2x (2cosx – 1) = 0

Either, sin2x = 0 = sin 0

Or, 2x = nπ

So, x = $\frac{{{\rm{n\pi }}}}{2}$.

Or, 2cosx – 1 = 0

Or, cosx = $\frac{1}{2}$

Or, cosx = cos $\frac{{\rm{\pi }}}{3}$.

So, x = 2nπ $ \pm $$\frac{{\rm{\pi }}}{3}$ = (6n$ \pm $1) $\frac{{\rm{\pi }}}{3}$

Hence, x = $\frac{{{\rm{n\pi }}}}{2}$, (6n$ \pm $1)$\frac{{\rm{\pi }}}{3}$, n ԑ Z.

 

c. cos3x – cosx = 0

Solution:

Here, cos3x – cosx = 0

Or, –2sin $\frac{{3{\rm{x}} + {\rm{x}}}}{2}$ . sin$\frac{{3{\rm{x}} - {\rm{x}}}}{2}$ = 0

Or, sin2x.sinx = 0

Either, sin2x.sinx = 0

Or, 2x = nπ.

So, x = $\frac{{{\rm{n\pi }}}}{2}$.

Or, sinx = 0 = sin0.

So, x = nπ.

Hence, x = nπ, $\frac{{{\rm{n\pi }}}}{2}$, n ԑ Z.

 

d. tan2x + tanx = 0

Solution:

Here, tan2x + tanx = 0

Or, $\frac{{{\rm{sin}}2{\rm{x}}}}{{{\rm{cos}}2{\rm{x}}}}$ + $\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$ = 0

Or, sin2x.cosx + cos2x.sinx = 0

Or, sin(2x + x) = 0

Or, sin3x = 0 = sin 0.

So. 3x = nπ + 0

So, x = ${\rm{n\pi }}$, $\frac{{{\rm{n\pi }}}}{2}$, n ԑ Z.

Since, $ - \frac{{\rm{\pi }}}{2}$ ≤ x ≤ $\frac{{\rm{\pi }}}{2}$, so,

i. For n = 0 àx = 0

ii. For n = –1 à x = $ - \frac{{\rm{\pi }}}{3}$

iii. For n = 1 à x = $\frac{{\rm{\pi }}}{3}$.

Hence, x = $ - \frac{{\rm{\pi }}}{3}$, 0, $\frac{{\rm{\pi }}}{3}$.

 

6. 

a. 2 cos²x + sinx.cosx – sin²x = 0

Solution:

Here, 2 cos²x + sinx.cosx – sin²x = 0.

Or, 2cos²x + 2sinx.cosx – sinx.cosx – sin²x = 0

Or, 2cosx(cosx + sinx) – sin(cosx + sinx) = 0

Or, (2cosx – sinx)(cosx + sinx) = 0

Either, 2cosx – sinx = 0

Or, sinx = 2cosx.

Or. $\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$ = 2

Or, tanx = tanα  [tanα = 2]

Or, x = nπ + α.

So, x = nπ + tan^–12.

Or, cosx + sinx = 0

Or, sinx = –cosx

Or. $\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$ = –1

Or, tanx = tan$\left( { - \frac{{\rm{\pi }}}{4}} \right)$

So, x = nπ + $\left( { - \frac{{\rm{\pi }}}{4}} \right)$

Hence, x = nπ – $\frac{{\rm{\pi }}}{4}$, nπ + tan^–12, n ԑ Z.

 

b. Sin2x – 4sinx – cosx + 2 = 0

Solution:

Sin2x – 4sinx – cosx + 2 = 0

Or, 2sinx.cosx – cosx – 4sinx + 2 = 0

Or, cosx(2sinx – 1) – 2(2sinx – 1) = 0

Or, (2sinx – 1)(cosx – 2) = 0

Either, 2sinx – 1 = 0

Or, sinx = $\frac{1}{2}$ = sin $\frac{{\rm{\pi }}}{6}$.

So, x = nπ + (–1)ⁿ$\frac{{\rm{\pi }}}{6}.$

Or, cosx – 2 = 0

Cos x = 2, is impossible.

Hence, x = nπ + (–1)ⁿ$\frac{{\rm{\pi }}}{6}$ , n ԑ Z.

 

7.

a. sin9θ = sinθ

Solution:

Here, sin9θ = sinθ

Or, sin9θ – sinθ = 0

Or, 2 cos$\frac{{9\theta  + \theta }}{2}$ .sin$\frac{{9\theta  - \theta }}{2}$ = 0

Or, cos5θ .sin4θ = 0

Either, sin4θ = 0 = sin 0.

So, 4θ = nπ

So, θ= $\frac{{{\rm{n\pi }}}}{4}$.

Or, cos5θ = 0 = cos $\frac{{\rm{\pi }}}{2}$.

Or, 5θ = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$

Or, θ= (4n$ \pm $1)$\frac{{\rm{\pi }}}{{10}}$.

Hence, θ = $\frac{{{\rm{n\pi }}}}{4}$, (4n$ \pm $1) $\frac{{\rm{\pi }}}{{10}}$, n ԑ Z.

 

b. tan5θ = cot2θ

Here,

tan5θ = cot2θ

or.tan5θ = tan $\left( {\frac{{\rm{\pi }}}{2} - 2\theta } \right)$

or, 5θ = nπ $ + \left( {\frac{{\rm{\pi }}}{2} - 2\theta } \right)$

or, 5θ + 2θ = (2n + 1) $\frac{{\rm{\pi }}}{2}$

So, θ= (2n + 1)${\rm{\: }}\frac{{\rm{\pi }}}{{14}}$, n ԑ Z.

 

c. tan2x – cotx = 0

Solution:

tan2x – cotx = 0

Or, tan 2x = tan $\left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)$

Or, 2x = nπ + $\left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)$

Or, 3x = (2n + 1) $\frac{{\rm{\pi }}}{2}$

So, x = (2n + 1)$\frac{{\rm{\pi }}}{6}$, n ԑ Z.

OR,

Tan2x – cotx = 9

Or, $\frac{{{\rm{sin}}2{\rm{\pi }}}}{{{\rm{cos}}2{\rm{x}}}}$ –$\frac{{{\rm{cosx}}}}{{{\rm{sinx}}}}$ = 0

Or, sin2x.sinx – cos2x.cosx = 0

Or, cos2x, cosx – sin2x.sinx = 0

Or, cos(2x + x) = 0

Or. Cos3x = cos $\frac{{\rm{\pi }}}{2}$

So, 3x = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$.

So, x = (4n$ \pm $1) $\frac{{\rm{\pi }}}{6}$, n ԑ Z.

 

d. tan mθ + cot nθ = 0

Solution:

tan mθ + cot nθ = 0

Or. $\frac{{\sin {\rm{m}}\theta }}{{\cos {\rm{m}}\theta }}$ + $\frac{{\cos {\rm{n}}\theta}}{{\sin {\rm{n}}\theta }}$ = 0.

Or, sin mθ. Sin nθ + cos mθ .cos nθ = 0

Or, cos (mθ – nθ) = 0

Or, cos (mθ – nθ) = cos $\frac{{\rm{\pi }}}{2}$

Or, (m – n)θ = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$.

So, θ = $\frac{{\left( {4{\rm{n}} \pm } \right){\rm{\pi }}}}{{2\left( {{\rm{m}} - {\rm{n}}}\right)}}$, n ԑ Z.

 

8.

a. cosθ + cos2θ + cos3θ = 0

Solution:

Here, cosθ + cos2θ + cos3θ = 0

Or, (cos3θ + cosθ) + cos2θ = 0.

Or, 2cos $\frac{{3\theta  + 2\theta }}{2}$ . cos$\frac{{3\theta  - 2\theta }}{2}$ + cos2θ = 0

Or, 2cos2θ.cosθ + cos2θ = 0

Or, cos θ(2cosθ + 1) = 0

Either, cos2θ = 0 = cos $\frac{{\rm{\pi }}}{2}$.

So, 2θ = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$

So, θ = (4n $ \pm $ 1) $\frac{{\rm{\pi }}}{4}$

Or, 2cosθ + 1 = 0.

Or, cosθ = $ - \frac{1}{2}$ = cos $\frac{{2{\rm{\pi }}}}{3}$

So, θ = 2nπ $ \pm $$\frac{{2{\rm{\pi }}}}{3}$ = (6n $ \pm $ 2) $\frac{{\rm{\pi }}}{3}$.

Hence, θ = (4n $ \pm $ 1) $\frac{{\rm{\pi }}}{4}$, (6n $ \pm $ 2) $\frac{{\rm{\pi }}}{3}$, n ԑ Z.

 

 

b. cosθ – sin3θ = cos2θ

Solution:

Here, cosθ – sin3θ = cos2θ.

Or, (cosθ – cos2θ) – sin3θ = 0

Or, 2 sin $\frac{{\theta  + 2\theta }}{2}$. Sin $\frac{{2\theta  - \theta }}{2}$– 2sin $\frac{{3\theta}}{2}$. Cos $\frac{{3\theta }}{2}$ = 0

Or. 2 sin $\frac{{3\theta }}{2}$$\left( {\sin \frac{\theta }{2} - \cos \frac{{3\theta }}{2}} \right)$ = 0

Either, sin $\frac{{3\theta }}{2}$ = 0 = sin 0.

So, $\frac{{3\theta }}{2}$ = nπ.

So, θ = $\frac{{2{\rm{n\pi }}}}{3}$.

Or, sin $\frac{\theta }{2}$ – cos $\frac{{3\theta }}{2}$ = 0

Or, cos $\frac{{3\theta }}{2}$ = sin $\frac{\theta }{2}$ = cos $\left( {\frac{{\rm{\pi }}}{2} - \frac{\theta }{2}} \right)$

So, $\frac{{3\theta }}{2}$ = 2nπ $ \pm $$\left( {\frac{{\rm{\pi }}}{2} - \frac{\theta }{2}} \right)$.

Or, $\frac{{3\theta }}{2}$ = 2nπ + $\frac{{\rm{\pi }}}{2}$ – $\frac{\theta }{2}$ and $\frac{{3\theta }}{2}$ = 2nπ – $\frac{{\rm{\pi }}}{2}$ + $\frac{\theta }{2}$.

Or, $\frac{{3\theta }}{2}$ + $\frac{\theta }{2}$ = 2nπ + $\frac{{\rm{\pi }}}{2}$ and $\frac{{3\theta }}{2}$ – $\frac{\theta }{2}$ = 2nπ – $\frac{{\rm{\pi }}}{2}$.

Hence, θ= $\frac{{2{\rm{n\pi }}}}{3}$, (4n + 1) $\frac{{\rm{\pi }}}{4}$, (4n – 1) $\frac{{\rm{\pi }}}{2}$, n ԑ Z.

 

c. tan θ + tan 2θ = tan 3θ

Solution:

Here, tan θ + tan 2θ = tan 3θ.

Or, tanθ + tan2θ – tan(θ + 2θ) = 0

Or, (tanθ + tan2θ) – $\frac{{{\rm{tan}}\theta  + {\rm{tan}}2\theta }}{{1 - {\rm{tan}}\theta .{\rm{tan}}2\theta }}{\rm{\: \: }}$= 0

Or, (tanθ + tan2θ) (1 – tanθ.tan2θ – 1) = 0

So, tanθ.tan2θ(tanθ + tan2θ) = 0

Either, tan θ = 0 = tan 0.

So, θ = nπ.

Or, tan 2θ = 0 = tan 0.

So, 2θ = nπ.

So, θ = $\frac{{{\rm{n\pi }}}}{2}$.

Or, tan θ + tan2θ = 0.

Or, tan θ = –tanθ = tan(–θ)

So, 2θ = nπ + (–θ)

SO, 2θ + θ = nπ

SO, θ = $\frac{{{\rm{n\pi }}}}{3}$.

Hence, θ = nπ, $\frac{{{\rm{n\pi }}}}{2}$, $\frac{{{\rm{n\pi }}}}{3}$, n ԑ Z.

 

d. 2sinx.sin3x = 1

Solution:

Here, 2sinx.sin3x = 1

Or, 2sinx(3sinx – 4sin³x) – 1 = 0.

Or, 6sin²x – 8sin⁴x– 1 = 0

Or, 8sin⁴x – 6sin²x + 1 = 0.

So, sin²x = $\frac{{6 \pm \sqrt {36 - 4.8.1} }}{{2.8}}$

Or, sin²x = $\frac{{6 \pm 2}}{{16}}$ = $\frac{{6 + 2}}{{16}}$. $\frac{{6 - 2}}{{16}}$

So, sin²x = $\frac{1}{2}$, $\frac{1}{4}$.

Now, sin² x = $\frac{1}{2}$ = ${\left( { \pm \frac{1}{{\sqrt 2 }}} \right)^2}$ = sin²$\frac{{\rm{\pi }}}{4}$.

So, x = nπ $ \pm $$\frac{{\rm{\pi }}}{4}$    [sin²x = sin² α à x = nπ*a]

Or, sin²x = $\frac{1}{4}$ = ${\left( { + \frac{1}{2}} \right)^2}$ = sin²$\frac{{\rm{\pi }}}{6}$.

So, x = nπ $ \pm $$\frac{{\rm{\pi }}}{6}$.

Here, x = nπ $ \pm $$\frac{{\rm{\pi }}}{6}$, nπ $ \pm $$\frac{{\rm{\pi }}}{4}$, n ԑ Z.

 

9.

a. cot²x – cosecx – 1 = 0

Solution:

Here, cot²x – cosecx – 1 = 0

Or, cosec²x – 1 – cosecx– 1 = 0

Or, cosec²x – cosecx – 2 = 0

Or, cosecx (cosecx – 2) + 1(cosecx – 2) = 0

Or, (cosec x – 2)(cosecx + 1) = 0

Either, cosec x – 2 = 0

Or, cosec x = 2.

Or, sinx = $\frac{1}{2}{\rm{\: }}$= sin $\frac{{\rm{\pi }}}{6}$.

So, x = nπ + (–1)ⁿ$\frac{{\rm{\pi }}}{6}$.

Or, cosecx + 1 = 0

Or, cosecx = –1

So, sinx = –1

So, x = (4n –1)$\frac{{\rm{\pi }}}{2}$

Hence, x = nπ + (–1)ⁿ. $\frac{{\rm{\pi }}}{6}$, (4n – 1)$\frac{{\rm{\pi }}}{2}$, n ԑ Z.

 

b. 2cosx + 1 = sinx

Solution:

Here, 2cosx + 1 = sinx

Squaring we have, 4cos²x + 4cosx + 1 = 1 – cos²x

Or, 5cos²x + 4cosx = 0

Or, cosx(5cosx + 4 ) =0

Either, cosx = 0

Or, cosx = cos $\frac{{\rm{\pi }}}{2}$.

So, x = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$ = $\left( {4{\rm{n}} \pm 1} \right)$$\frac{{\rm{\pi }}}{2}$.

Or, 5cosx + 4 = 0

Or, cos x = $ - \frac{4}{5}$.

Or. Cosx = cosα [cosα = $ - \frac{4}{5}$]

Or, x = 2nπ $ \pm $ α

Hence, x = 2nπ $ \pm $ cos^–1$\left( { - \frac{4}{5}} \right)$, (4n$ \pm $1) $\frac{{\rm{\pi }}}{2}$.

 

c. secx . tanx = $\sqrt 2 $

Soln;

here, secx . tanx = $\sqrt 2 $

Or, sec²x. tan²x = 2 [squaring on both side.]

Or, (1 + tan²x)tan²x = 2

Or, tan⁴x + tan²x – 2 = 0

Or, tan⁴ x + 2tan²x – tan²x – 2 = 0

Or, tan²x (tan²x + 2) – 1(tan²x + 2) = 0

Or, (tan²x + 2)(tan²x – 1) = 0

Either, tan²x – 1 = 0

Or, tan²x = 1 = ${\left( { \pm 1} \right)^2}$ = tan²$\left( {\frac{{\rm{\pi }}}{4}} \right)$^.

So, x = nπ $ \pm $$\frac{{\rm{\pi }}}{4}$.

Or, tan²x + 2 = 0

So, tan² = –2 [is imaginary]

Hence, x = nπ $ \pm $$\frac{{\rm{\pi }}}{4}$, n ԑ Z.

 

d. 4sin⁴x – cos²2x = 0

Solution:

Here, 4sin⁴x – cos²2x = 0

Or, (2sin²x)² – cos²2x = 0

Or, (1 – cos2x)² – cos²2x = 0

Or, 1 – 2cos2x + cos²2x – cos²2x = 0

Or, 1 – 2cos2x = 0

Or, cos2x = $\frac{1}{2}$ = cos $\frac{{\rm{\pi }}}{3}{\rm{\: }}$

Or, 2x = 2nπ $ \pm $$\frac{{\rm{\pi }}}{3}$.

Hence, x = (6n$ \pm $1) $\frac{{\rm{\pi }}}{6}$ , n ԑ Z.

 

10.

a. cos θ + cos 3θ + cos 5θ + cos 7θ = 0

Solution:

cos θ + cos 3θ + cos 5θ + cos 7θ = 0

(cos 7θ + cos θ) + (cos 5θ + cos 3θ) = 0

Or, 2 cos $\frac{{7\theta  + \theta }}{2}$, cos $\frac{{7\theta  - \theta }}{2}$ + 2cos$\frac{{5\theta + 3\theta }}{2}$ , cos $\frac{{5\theta  - 2\theta }}{2}$ = 0

Or, 2cos4θ.cos3θ + 2cos4θ.cosθ = 0

Or, cos4θ(cos3θ + cosθ) = 0

Or, cos 4θ. 2cos $\frac{{3\theta  + \theta }}{2}$.cos $\frac{{3\theta  - \theta }}{2}$ = 0

So, cos 4θ. Cos 2θ. Cos θ= 0

Either, cos 4θ = 0 = cos $\frac{{\rm{\pi }}}{2}$.

So, 4θ = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$

SO, θ= (4n$ \pm $1) $\frac{{\rm{\pi }}}{8}$.

Or, cos 2θ = 0 = cos $\frac{{\rm{\pi }}}{2}$.

So, 2θ = 2nπ $ \pm {\rm{\: }}$$\frac{{\rm{\pi }}}{2}$

So, θ = (4n$ \pm $1) $\frac{{\rm{\pi }}}{4}$

Or, cosθ= 0 = cos$\frac{{\rm{\pi }}}{2}$

So, θ = 2nπ $ \pm $$\frac{{\rm{\pi }}}{2}$.

So, θ = (4n$ \pm $1) $\frac{{\rm{\pi }}}{2}$.

Hence, θ = (4n$ \pm $1) $\frac{{\rm{\pi }}}{2}$, (4n$ \pm $1) $\frac{{\rm{\pi }}}{4}$, (4n$ \pm $1) $\frac{{\rm{\pi }}}{8}$, n ԑ Z.

 

b. tanθ + tan2θ + tan3θ = 0

Solution:

Here, tanθ + tan2θ + tan3θ = 0

Or, tan θ + tan2θ + tan(θ+2θ) = 0

Or, (tanθ + tan2θ) + $\frac{{{\rm{tan}}\theta  + {\rm{tan}}2\theta }}{{1 - {\rm{tan}}\theta .{\rm{tan}}2\theta }}$ = 0

Or, (tanθ + tan2θ)(1 – tanθ.tan2θ + 1) = 0

Or, (tanθ + tan2θ)(2 – tanθ.tan2θ) = 0

Either, tanθ + tan2θ = 0

Or, tan 2θ= –tanθ = tan(–θ)

So, 2θ = nπ + (–θ)

Or, 3θ= nπ

SO, θ= $\frac{{{\rm{n\pi }}}}{3}$

Or, 2 – tanθ, tan2θ = 0

Or, 2 – tan θ. $\frac{{2{\rm{tan}}\theta }}{{1 - {{\tan }^2}\theta }}$ = 0

Or, 2 – 2tan² θ – 2tan² θ = 0

Or, 2(1 – 2tan²θ) = 0

Or, 1 – 2tan²θ = 0

Or, tan²θ= $\frac{1}{2}$

Or, tan²θ = tan²α

Or, tan²θ = tan²α      [tan²α = $\frac{1}{2}$]

SO, θ= nπ $ \pm $ α.

So, θ = nπ $ \pm {\rm{\: }}$tan^–1$\left( {\frac{1}{{\sqrt 2 }}} \right)$    [tan²α = $\frac{1}{2}$]

And θ = $\frac{{{\rm{n\pi }}}}{3}$, n ԑ Z.

 

11. Find all the solution Tanθ – 3cotθ = 2tan3θ that lie between 0⁰ and 360⁰.

Solution:

Here the given equation is:

Tanθ – 3cotθ = 2tan3θ

Or, tanθ – $\frac{3}{{{\rm{tan}}\theta }}$ = $\frac{{2\left( {3{\rm{tan}}\theta  - {{\tan }^3}\theta } \right)}}{{1 - 3{{\tan }^2}\theta }}$

Or, (tan²θ – 3)(1 – 3tan² θ) = 2tanθ (3tanθ – tan³θ)

Or, tan²θ – 3tan⁴θ – 3 + 9tan²θ = 6tan²θ – 2tan⁴θ.

Or, –tan⁴θ + 4tan²θ–3 = 0

Or, tan⁴θ – 4tan²θ + 3 = 0

Or, tan⁴θ – 3tan²θ – tan² θ + 3 = 0

Or, tan²θ(tan²θ – 3) – 1(tan²θ – 3)$.$

Or, (tan²θ – 3)(tan²θ – 1).

Either, tan³θ – 3 = 0.

Or, tanθ = $ \pm \sqrt 3 $

So, θ = 60°,24–°,120°,300°

Or, tan²θ – 1 = 0

Or, tanθ= $ \pm $1

So, θ= 45°,135°,225°,315°.
Hence, θ= 45°,60°,120°,135°,225°,240°,300°,315°.

I.e. θ= $\frac{{\rm{\pi }}}{4}$, $\frac{{\rm{\pi }}}{3},\frac{{2{\rm{\pi }}}}{3},\frac{{3{\rm{\pi }}}}{3},\frac{{5{\rm{\pi }}}}{4},\frac{{4{\rm{\pi }}}}{3},\frac{{5{\rm{\pi }}}}{3},\frac{{7{\rm{\pi }}}}{4}$.

 

12. Find the solution of the equation (General solution not required)

Solution:

Here, given equations are:

tanx + tany = 2…(i)

And 2 cosx. Cosy = 1 …(ii)

Now, from (i), we have, $\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$ + $\frac{{{\rm{siny}}}}{{{\rm{cosy}}}}$ = 2

Or, sinxcosy + cosx. Siny = 2 cosx . cosy.

Or, sin(x + y) = 1   [from(2)]

Or, sin(x+y) = sin $\frac{{\rm{\pi }}}{2}$.

SO, x + y = $\frac{{\rm{\pi }}}{2}$ ….(3)

Again from (2) we have,

2cosx.cosy = 1

Or, cos(x+y) + cos(x – y) = cos 0

Or, cos $\frac{{\rm{\pi }}}{2}$ + cos(x –y) = cos 0[from(3)]

Or, 0 + cos(x – y) = cos 0.

So, x – y = 0…(4)

Now, adding (3) and (4), we have,

2x = $\frac{{\rm{\pi }}}{2}$

So, x = $\frac{{\rm{\pi }}}{4}$.

So, from (4), y = $\frac{{\rm{\pi }}}{4}$.

Hence, x = $\frac{{\rm{\pi }}}{4}$ , y = $\frac{{\rm{\pi }}}{4}$.

 

13.

(i) tanθ + tan2θ + tan3θ = tanθ.tan2θ.tan3θ

Solution:

tanθ + tan2θ + tan3θ = tanθ.tan2θ.tan3θ

Or, (tanθ+tan2θ) + tan3θ(1– tanθ.tan2θ) = 0

Or, $\frac{{{\rm{tan}}\theta  + {\rm{tan}}2\theta }}{{1 - {\rm{tan}}\theta .{\rm{tan}}2\theta }}$ + tan3θ = 0

Or, tan (θ + 2θ) + tan3θ = 0    [formula of tan(A+B)]

Or, tan3θ + tan3θ = 0

Or, 2tan3θ = 0

Or, tan3θ = tan0

So, 3θ = nπ + 0

So, θ = $\frac{{{\rm{n\pi }}}}{3}$, n ԑ Z.

 

(ii) tanθ + tan2θ + tanθ.tan2θ = 1

Solution:

tanθ + tan2θ + tanθ.tan2θ = 1

Or, tanθ + tan2θ = 1 – tanθ.tan2θ

Or, $\frac{{{\rm{tan}}\theta  + {\rm{tan}}2\theta }}{{1 - {\rm{tan}}\theta .{\rm{tan}}2\theta }}$ = 1

Or,tan(θ+2θ) = tan $\frac{{\rm{\pi }}}{4}$

Or. Tan 3θ = tan $\frac{{\rm{\pi }}}{4}$

Or, 3θ = nπ + $\frac{{\rm{\pi }}}{4}$.

So, θ= (4n + 1) $\frac{{\rm{\pi }}}{{12}}$, n ԑ Z.

 

(iii) tan$\left( {\frac{{\rm{\pi }}}{4} + \theta } \right)$ + tan $\left( {\frac{{\rm{\pi }}}{4} - \theta } \right)$ = 4

Solution:

tan$\left( {\frac{{\rm{\pi }}}{4} + \theta } \right)$ + tan $\left( {\frac{{\rm{\pi }}}{4} - \theta } \right)$ = 4.

Or, $\frac{{\tan \frac{{\rm{\pi }}}{4}{\rm{\: }} + \tan \theta }}{{1 - \tan \frac{{\rm{\pi }}}{4}.{\rm{tan}}\theta }}$ + $\frac{{\tan \frac{{\rm{\pi }}}{4} - \tan \theta }}{{1 + \tan \frac{{\rm{\pi }}}{4}.{\rm{tan}}\theta }}$ = 4

Or, $\frac{{1 + {\rm{tan}}\theta }}{{1 - {\rm{tan}}\theta }}$ + $\frac{{1 - {\rm{tan}}\theta }}{{1 + {\rm{tan}}\theta }}$ = 4

Or, (1+tanθ)² + (1 – tanθ)² = 4(1 – tanθ)(1 + tanθ)

or, 1 + 2tanθ + tan²θ + 1 – 2 tanθ + tan²θ = 4 – 4 tan²θ

or, 6tan²θ = 2

or, tan²θ = $\frac{1}{3}$ = tan²$\left( {\frac{{\rm{\pi }}}{6}} \right)$

So, θ = nπ $ \pm $$\frac{{\rm{\pi }}}{6}$, n ԑ Z.

 

14.

(i) Sin2x tanx + 1 = sin2x + tanx

Solution:  

Given equation is:

Sin2x tanx + 1 = sin2x + tanx

Or, sin2x tanx – sin2x – tanx + 1 = 0

Or, sin2x(tanx – 1) – 1(tanx – 1) = 0

Or, (tanx – 1)(sin2x – 1) = 0

Either, tanx – 1 =0

Or, tanx = 1 = tan $\frac{{\rm{\pi }}}{4}$.

So, x = nπ + $\frac{{\rm{\pi }}}{4}$ = (4n + 1)$\frac{{\rm{\pi }}}{4}$.

OR, sin2x – 1 = 0

Or, sin2x = 1

So, 2x = (4n + 1)  $\frac{{\rm{\pi }}}{2}$

SO, x = (4n + 1) $\frac{{\rm{\pi }}}{4}$

Hence, x = (4n + 1) $\frac{{\rm{\pi }}}{4}$, n ԑ Z.

 

(ii) 2sinxtanx + 1 = tanx + 2sinx

Solution:

Given equation is:

2sinxtanx + 1 = tanx + 2sinx.

Or. 2sinx tanx – tanx – 2sinx + 1 = 0

Or, tanx(2sinx – 1) – 1 (2sinx – 1) = 0

Or, (2sinx – 1)(tanx – 1) = 0

Either, tanx – 1 = 0

Or, tan x = 1 = tan $\frac{{\rm{\pi }}}{4}$.

So, x = nπ + $\frac{{\rm{\pi }}}{4}$ = (4n + 1) $\frac{{\rm{\pi }}}{4}$.

Or, 2sinx – 1 = 0

Or, sinx = $\frac{1}{2}$ = sin $\frac{{\rm{\pi }}}{6}$

So, x = nπ+ (–1)ⁿ$\frac{{\rm{\pi }}}{6}$.

Hence, x = (4n + 1) $\frac{{\rm{\pi }}}{4}$, nπ+ (–1)ⁿ$\frac{{\rm{\pi }}}{6}$., n ԑ Z.

 

15. 

(i) Cos2x + sin2x = cosx + sinx

Solution:

Here given equation is:

Cos2x + sin2x = cosx + sinx

Or, $\frac{1}{{\sqrt 2 }}$cos2x + $\frac{1}{{\sqrt 2 }}$sin2x = $\frac{1}{{\sqrt 2 }}$cosx + $\frac{1}{{\sqrt 2 }}$sinx.

Or, cos $\frac{{\rm{\pi }}}{4}$cos2x + sin $\frac{{\rm{\pi }}}{4}$ sin2x = cos $\frac{{\rm{\pi }}}{4}$ cosx + sin $\frac{{\rm{\pi }}}{4}$ sinx

Or, cos $\left( {2{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)$ = cos $\left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)$

SO, 2x – $\frac{{\rm{\pi }}}{4}$ = 2nπ $ \pm $$\left( {{\rm{x}} - \frac{{\rm{\pi }}}{4}} \right)$

Or, $\{ \begin{array}{*{20}{c}}{2{\rm{x}} - \frac{{\rm{\pi }}}{4} = 2{\rm{n\pi }} + {\rm{x}} - \frac{{\rm{\pi }}}{4}{\rm{\: \: }}}\\{2{\rm{x}} - \frac{{\rm{\pi }}}{4} = 2{\rm{\pi n}} - {\rm{x}} + \frac{{\rm{\pi }}}{4}{\rm{\: }}}\end{array}$

Or, $\{ \begin{array}{*{20}{c}}{{\rm{x}} = 2{\rm{n\pi \: \: }}}\\{3{\rm{x}} = 2{\rm{n\pi }} + \frac{{\rm{\pi }}}{2}{\rm{\: }}}\end{array}$

So, x = 2nπ, (4n + 1) $\frac{{\rm{\pi }}}{6}$, n ԑ Z.

 

(ii) Cosx – sinx = cosα + sinα

Solution:

Given equation is:

Cosx – sinx = cosα + sinα.

Or, $\frac{1}{{\sqrt 2 }}$cosx  –$\frac{1}{{\sqrt 2 }}$sinx = $\frac{1}{{\sqrt 2 }}$cosα + $\frac{1}{{\sqrt 2 }}$sinα.

Or, cos $\frac{{\rm{\pi }}}{4}$cosx – sin$\frac{{\rm{\pi }}}{4}$sinx = cos$\frac{{\rm{\pi }}}{4}$cosα  + sin$\frac{{\rm{\pi }}}{4}$.sinα.

Or, cos$\left( {{\rm{x}} + \frac{{\rm{\pi }}}{4}} \right)$ = cos$\left( {\alpha  + \frac{{\rm{\pi }}}{4}} \right)$

SO, x + $\frac{{\rm{\pi }}}{4}$ = 2nπ $ \pm $$\left( {\alpha  - \frac{{\rm{\pi }}}{4}} \right)$

Thus,

Or, $\{ \begin{array}{*{20}{c}}{{\rm{x}} + \frac{{\rm{\pi }}}{4} = 2{\rm{\pi n}} + \alpha  - \frac{{\rm{\pi }}}{4}}\\{{\rm{x}} + \frac{{\rm{\pi }}}{4} = 2{\rm{n\pi }} - \alpha  + \frac{{\rm{\pi }}}{4}}\end{array}$

Or, $\{ \begin{array}{*{20}{c}}{{\rm{x}} = 2{\rm{n\pi }} + \alpha  - \frac{{\rm{\pi }}}{2}}\\{{\rm{x}} = 2{\rm{n\pi }} - \alpha {\rm{\: }}}\end{array}$

Hence, ${\rm{x}} = 2{\rm{n\pi }} - \alpha $, 2nπ – $\frac{{\rm{\pi }}}{2}$ + α, n ԑ Z.