KC Sinha Mathematics Solution Class 11 Chapter 7 त्रिकोणमिति समीकरण (Trigonometric equations) Exercise 7.1

Exercise 7.1

Question 1

Find the principal solution of the following trigonometric equations :
(i) $\sin x=\dfrac{1}{2}$
Sol :

$\sin x=\frac{1}{2}$

$\sin x=\sin \frac{\pi}{6}$

[$\sin \theta=\sin \alpha \Rightarrow$
$\theta=n \pi+(-1)^{n} \alpha, n \in z$]

$x=n \pi+(-1)^{n} \frac{\pi}{6}, n \in z$

$0 \leq x<\frac{3 60^{\circ}}{2}$


Principal solution :

when n=0 , $x=\frac{\pi}{6}$

when n=1 , $x=\pi - \frac{\pi}{6}$


(ii) tan x=-1

Sol :
$\tan x=-\tan \frac{\pi}{4}$

$\tan x=\tan \left(\pi-\frac{\pi}{4}\right)$

$\tan x=\tan \frac{3 \pi}{4}$

[when $\tan \theta=\tan \alpha$ , then

$\theta=n \pi+\alpha, n \in z$]


$x=n \pi+\frac{3 \pi}{4}, n \in z$

Principal value :

$x=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$

$x=2 \pi-\frac{\pi}{4}= \frac{7 \pi}{4}$


(iii) $\cot x=\frac{1}{\sqrt{3}}$
Sol :




(iv) $\sec x=-\frac{2}{\sqrt{3}}$

Question 2

(i) $\cos \theta=\frac{1}{2}$
Sol :
$\cos \theta=\frac{1}{2}$

$\cos \theta=\cos \frac{\pi}{3}$

[if $\cos \theta=\cos \alpha$ then  ,

$\theta=2 n \pi \pm \alpha, n \in z$ ]


$\theta=2 n \pi \pm \frac{\pi}{3}, n \in z$


(ii) $\sin \theta=-1$
Sol :
$\sin \theta=\sin \left(-\frac{\pi}{2}\right)$

[if $\sin \theta=\sin \alpha$ then ,

$\theta=n \pi+(-1)^{n}, \alpha n\in z$ ]

$\theta=n \pi+(-1)^{n} \cdot\left(\frac{-\pi}{2}\right), n \in z$

$\theta=n \pi+(-1)^{n+1} \cdot\left(\frac{\pi}{2}\right) \cdot n \in z$


(iii) $3 \tan ^{2} \theta=1$
Sol :
$\tan ^{2} \theta=\frac{1}{3}$

$\tan ^{2} \theta=\left(\frac{1}{\sqrt{3}}\right)^{2}$

$\tan ^{2} \theta=\tan ^{2} \frac{\pi}{6}$

[if  $\tan ^{2} \theta=\tan ^{2} \alpha$ then ,
$\theta=n \pi \pm \alpha, n \in z$ ]

$\theta=n \pi \pm \frac{\pi}{6}, n \in z$



(iv) $2 \sin \theta=\sqrt{3}$

(v) $\sin x=-\frac{\sqrt{3}}{2}$
Sol :
$\sin x=-\sin \frac{\pi}{3}$

$\sin x=\sin \left(\pi+\frac{\pi}{3}\right)$


$\sin x=\sin \frac{4 \pi}{3}$

$x=n \pi+(-1)^{n} \frac{4 \pi}{3}, n \in z$

Question 3

Sol :
$\sin  \theta=1$

$\sin \theta=\sin \frac{\pi}{2}$

$\theta=\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9\pi}{2}$

or

$-\frac{3 \pi}{2},-\frac{7 \pi}{2}, \frac{-11\pi}{2}$

$\therefore \quad \theta=(4 n+1) \frac{\pi}{2}, n \in z$

Question 4

(i) cosmx+cosnx=0
Sol :
$\cos C+\cos D=2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$

2 \cos \frac{m x+n x}{2} \cos \frac{m x-n x}{2}=0

$\cos \frac{(m+n) x}{2} \cos \frac{(m-n) x}{2}=0$


$\cos \left(\frac{m+n}{2}\right) x=0$

$\frac{(m+n) x}{2}=(2 r+1) \frac{\pi}{2}, r \in 2$

$x=\frac{(2 x+1)}{m+n} \pi, r \in z$


$\cos \left(\frac{m-n}{2}\right) x=0$

$\frac{(m-n)}{2} x=(2 r+1) \frac{\pi}{2}, r \in z$

$x=\frac{(2 r+1) \pi}{m-n}, r \in z$


$\therefore x=\frac{(2 r+1) \pi}{m \pm n}, r \in z$




(ii) tanx.tan4x=1
Sol :

0=1-tanx.tan4x

$1-\frac{\sin x \sin 4 x}{\cos x \cos 4 x}=0$

$\frac{\cos x \cos 4 x-\sin x \sin 4 x}{\cos x \cos 4 x}=0$

$\frac{\cos (x+4x)}{\cos x \cos 4 x}=0$

cos5x=0

$5 x=(2 n+1) \frac{\pi}{2}, n \in z$

$x=(2 n+1) \frac{\pi}{10}, n \in z$


(iii) tan5θ=cot2θ
Sol :

0=cot2θ-tan5θ

$0=\frac{\cos 2 \theta}{\sin 2 \theta}-\frac{\sin 5 \theta}{\cos 5 \theta}$

$0=\frac{\cos 5 \theta \cos \theta-\sin \theta \sin 2 \theta}{\sin 2 \theta \cos 5 \theta}$

$0=\frac{\cos (5 \theta+2 \theta)}{\sin 2 \theta \cos 5 \theta}$

0=cos7θ

$7 \theta=(2 n+1) \frac{\pi}{2}, n \in I$

$\theta=(2 n+1) \frac{\pi}{14},n\in I$


(iv) $\tan x=-\tan \left(x+\frac{\pi}{3}\right)$
Sol :
$\tan x=\tan \left[2 \pi-\left(x+\frac{\pi}{3}\right)\right]$

$\tan x=\tan \left(\frac{5 \pi}{3}-x\right)$

$x=n \pi+\frac{5 \pi}{3}-x, n \in z$

$2 x=n \pi+\frac{5 \pi}{3}, n \in z$

$x=\frac{n \pi}{2}+\frac{5 x}{6}, n \in z$

Question 5

Sol :
$\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$

$\tan (\pi \cos \theta)=\tan \left[\frac{\pi}{2}-\pi \sin \theta\right]$

$\pi \cos \theta=\frac{\pi}{2}-\pi \sin \theta$

$\pi \sin \theta+\pi \cos \theta=\frac{\pi}{2}$

$\pi(\sin \theta+\cos \theta)=\frac{\pi}{2}$

$\sin \theta+\cos \theta=\frac{1}{2}$

Question 6

(i) $\sin x+\cos x=\sqrt{2}$
Sol :
$\frac{\sin x+\cos x}{\sqrt{2}}=1$

$\sin x \cdot \frac{1}{\sqrt{2}}+\cos x \cdot \frac{1}{\sqrt{2}}=1$

$\sin x \cdot \frac{1}{\sqrt{2}}+\cos x \cdot \frac{1}{\sqrt{2}}=1$

$\sin x+\frac{\pi}{4}+\cos x \cos \frac{\pi}{4}=1$

$\cos \left(x-\frac{\pi}{4}\right)=\cos 0$

$x-\frac{\pi}{4}=2 n \pi \pm 0, n \in z$

$a-\frac{\pi}{4}=2 n \pi , n \in z$

$x=2 n x+\frac{\pi}{4}, n \in z$

$x=\frac{8 n \pi+\pi}{4}, n \in z$

$x=(8 n+1) \frac{\pi}{4}, n\in z$

Question 7

$2 \cos x+\sin x=1$
Sol :
[] $r \sin \alpha=2, r \cos \alpha=1$

[]

$r^{2} \sin ^{2} \alpha+r^{2} \cos ^{2} \alpha=2^{2}+1^{2}$

$r^{2}\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)=5$

$r=\sqrt{5}$

$r \sin \alpha \cos x+r \cos \alpha \sin x=1$

$r(\sin \alpha \cos x+\cos \alpha \sin x)=1$

$\sqrt{5} \sin(x+\alpha)=1$

$\sin (x+\alpha)=\frac{1}{\sqrt{5}}$

sin(x+ɑ)=cosɑ

$\sin (x+\alpha)=\sin \left(\frac{\pi}{2}-\alpha\right)$

$x+\alpha=n \pi+(-1)^{n} \cdot\left(\frac{\pi}{2}-\alpha\right), n \in z$

[]

$x+\alpha=n x+\frac{\pi}{2}-\alpha, n \in z_{1}$

$x=\frac{2 n \pi+\pi}{2}-2 \alpha, n \in z_1$

$x=(2 n+1) \frac{\pi}{2}-2 \alpha, n \in z_1$

[]

$\cos \alpha=\frac{1}{\sqrt{5}}$


$x+\alpha=n x-\left(\frac{\pi}{2}-\alpha\right), n \in z_{2}$

$x+\alpha=n \pi-\frac{\pi}{2}+\alpha, n \in z_{2}$

$x=\frac{2 n \pi-\pi}{2}, n \in z_{2}$

$x=(2 n-1) \frac{\pi}{2}, n \in z_2$

Question 8

Sol :
$\cos 2 \theta-\sin 2 \theta=\cos \theta-\sin \theta-1$

$2 \cos ^{2} \theta-1-2 \sin \theta \cos \theta=\cos \theta-\sin \theta-1$

$2 \cos \theta(\cos \theta-\sin \theta)=\cos \theta-\sin \theta$

$2 \cos \theta =1$

$\cos \theta =\frac{1}{2}$


$\cos \theta= \cos \frac{\pi}{3}$

$\theta=2 n \pi \pm \frac{\pi}{3}, n \in z$

$\theta=\frac{5 \pi}{3}, \frac{7 \pi}{3}, \frac{8 \pi}{3}$

$\theta=300^{\circ},420^{\circ},480^{\circ}$

Question 9

(i) sinx+cosx+secx+cosecx=0
Sol :
$\sin x+\cos x+\frac{1}{\cos x}+\frac{1}{\sin x}=0$

$\sin x+\cos x+\frac{\sin x+\cos x}{\cos x \sin x}=0$

$\left(\sin x+\cos x\left(1+\frac{1}{\cos x \sin x}\right)=0\right.$

sinx+cosx=0

sinx=-cosx

$\frac{\sin x}{\cos x}=-1$

tanx=-1

$\tan x=-\tan \frac{\pi}{4}$

$x=n \pi+\left(-\frac{\pi}{4}\right) \quad n \in z$

$x=n \pi-\frac{\pi}{4}, n \in z$


(ii) cos3θsin³θ+sin3θcos³θ=0
Sol :
$\left(4 \cos ^{3} \theta-3 \cos \theta\right) \sin^{3} \theta+\left(3 \sin \theta-4 \sin ^{3} \theta\right) \cos ^{3} \theta=0$

$4 \cos^{3} \theta -\sin ^{3} \theta-3 \cos \sin ^{3} \theta+3 \sin \theta \cos ^{3} \theta-4 \sin ^{3} \theta \cos ^{3} \theta=0$

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

sinθcosθcos2θ=0

$\frac{1}{2}(2 \sin \theta \cos \theta) \cos 2 \theta=0$

$\frac{1}{2} \sin 2 \theta \cos 2 \theta=0$

sin2θcos2θ=0

$\frac{1}{2}(2 \sin 2 \theta \cos 2 \theta)=0$

sin2(2θ)=0

sin4θ=0

$4 \theta=n \pi, n \in z$

$\theta=\frac{n \pi}{4}, n\in z$

Question 10

(i) tanθ+tan2θ+√3tanθ.tan2θ=√3
Sol :
tanθ+tan2θ=√3-√3tanθtan2θ

tanθ+tan2θ=√3(1-tanθ.tan2θ)

$\frac{\tan \theta+\tan 2 \theta}{1-\tan \theta \tan 2 \theta}=\sqrt{3}$

tan(θ+2θ)=√3

$\tan (\theta+2 \theta)=\sqrt{3}$

$\tan 3 \theta=\tan \frac{\pi}{3}$

$3 \theta=n \pi+\frac{\pi}{3}, n \in z$

$\theta=\frac{n \pi}{3}+\frac{\pi}{9} \quad, n \in z$

$\theta=\frac{3 n \pi+\pi}{9}, n \in z$

$\theta=(3 n+1) \frac{\pi}{9}, n\in z$



(ii) tanx+tan2x+tanx.tan2x=1
Sol :
tanx+tan2x=1-tanx.tan2x

$\frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}=$

$\tan 3 x=\tan \frac{\pi}{4}$

$3 x=n \pi+\frac{\pi}{4} \quad, n \in z$

$x=\frac{n \pi}{3}+\frac{\pi}{12}, n \in z$

$x=\frac{4 n \pi+\pi}{12}, n \in z$

$=(4 n+1) \frac{\pi}{12}, n \in z$


(iii) tanx+tan4x+tan7x=tanx.tan4x.tan7x
Sol :
tanx+tan2x=-tan3x+tanx.tan2x.tan3x

tanx+tan2x=-tan3x(1-tanx.tan2x)

$\frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}=-\tan 3 x$

tan3x=-tan3x

2tan3x=0

tan3x=0

3x=n𝜋 , n𝜀z

$x=\frac{n \pi}{3}, n \in z$


(iv) tanx+tan2x+tan3x=tanx.tan2x.tan3x
Sol :

Question 11

(i) 2sin²x+sinx=3 , (0°≤x≤1000°)
Sol :
2sin²x+sinx-3=0

2sin²x+3sinx-2sinx-3=0

sinx(2sinx+3)+(2sinx+3)=0

(2sinx+3)(sinx-1)

CASE-I
2sinx+3=0

$\sin x=-\frac{3}{2}$

Not possible because

-1≤sinx≤-1


CASE-II
sinx-1=0

six=1

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

$x=n \pi+(-1)^{n} \frac{\pi}{2}, n \in$

x=90° , 450° , 810°


(ii) 2cos²x-cosx=3 , (0°≤x≤1000°)
Sol :
2cos²x-cosx-3=0

2cos²x-3cosx+2cosx-3=0

cosx(2cosx-3)+1(2cosx-3)=0

(2cosx-3)(cosx+1)=0

CASE-I
2cosx-3=0

$\cos x=\frac{3}{2}$

Not possible because

-1≤cosx≤1


CASE-II
cosx+1=0

cosx=-1

cosx=cos𝜋

[if cosθ=cos𝛼

then θ=2n𝜋± 𝛼 , n 𝝐z]

x=2n𝜋± 𝛼 , n 𝝐z

x=180° , 540° , 900°

(iii) 2sin²x+sin²x2x=2
Sol :
2sin²x+(2sinxcosx)²=2

2sin²x+4sin²xcos²x=2

2(sin²x+2sin²xcos²x)=2

sin²x+2sin²x(1-sin²x)=1

sin²x+2sin²x-2sin⁴x=1

3sin²x-2sin⁴x=1

0=2sin⁴x-3sin²x+1

2sin⁴x-2sin²x-sin²x+1=0

2sin²x(sin²x-1)-1(sin²x-1)=0

(sin²x-1)(2sin²x-1)=0

CASE-I
sin²x-1=0
sin²x=1²

$\sin ^2x=\sin ^{2} \frac{\pi}{2}$

$x=n \pi \pm \frac{\pi}{2}, n \in z$

$x=\frac{2 n \pi \pm \pi}{2}, n \in z$

$x=(2 n \pm 1) \frac{\pi}{2}, n \in z$


CASE-II
2sin²x-1=0
2sin²x=1
$\sin ^{2} x=\left(\frac{1}{\sqrt{2}}\right)^{2}$
$\sin^{2} x=\sin ^{2} \frac{\pi}{4}$
$x=n \pi \pm \frac{\pi}{4}, n \in z$


(iv) 2cos²x+3sinx=0

Question 12

(i) tan²x+cot²x=2
Sol :
sec²-1+cosec²x-1=2
sec²x+cosec²x=4

$\frac{1}{\cos ^{2} x}+\frac{1}{\sin^{2} x}=4$

$\frac{\sin ^{2} x+\cos ^{2} x}{\cos ^{2} x \sin ^{2} x}=4$

1=4cos²x.sin²x
1=(2sinx.cosx)²
1=sin²2x
$\sin ^{2} \frac{\pi}{2}=\sin^{2} 2 x$

$2 x=n x \pm \frac{\pi}{2}, n \in z$

$2 x=\frac{2 n\pi \pm \pi}{2}, n \in z$

$2 x=(2 n \pm 1) \frac{\pi}{2}, n \in z$

$x=(2 n \pm 1) \frac{\pi}{4}, n\in z$


(ii) tan²x=3cosec²x-1
Sol :
sec²x-y=3cosec²x-1

$\frac{1}{\cos ^{2} x}=\frac{3}{\sin ^{2} x}$

$\frac{\sin ^{2} x}{\cos ^{2} x}=3$

tan²x=(√3)²

$\tan ^{2} x=\tan ^{2}\left(\frac{\pi}{3}\right)$

$x=n \pi \pm \frac{\pi}{3}, n \in z$

Question 13

(i) cos6x+cos4x=sin3x+sinx
Sol :
$\cos C+\cos D=2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$

$\sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$

$2 \cos \frac{6 x+4 x}{2} \cos \frac{6 x-4 x}{2}=2 \sin 2 \frac{3 x+x}{2} \cos \frac{3 x-x}{2}$

cos5x.cosx=sin2xcosx

cos5x=sin2x

$\cos 5 x=\cos \left(\frac{\pi}{2}-2 x\right)$

$5 x=2 n \pi \pm\left(\frac{\pi}{2}-2 x\right) , n \in z$

$7 x=\frac{4 n \pi+\pi}{2}, n \in z$

$7 x=(4 n+1) \frac{\pi}{2}, n+z$



CASE-I
$5 x=2 n \pi+\left(\frac{\pi}{2}-2 x\right) , n\in z$

$5x=2 n \pi+\frac{\pi}{2}-2 x$

$x=(4 n+1) \frac{\pi}{14}, n \in z$

CASE-II
$5 x=2 n \pi-\left(\frac{\pi}{2}-2 x\right) , n\in z$

$5 x=2 n \pi-\frac{\pi}{2}+2 x , n\in z$

$3 x=2 n \pi-\frac{\pi}{2}, n \in z$

$3 x=\frac{4 n \pi-\pi}{2}, n \in z$

$3 x=(4n-1) \frac{\pi}{2}, n\in 2$

$x=(4n -1) \frac{\pi}{6}, n \in z$


(ii) cos5x+cos3x+cosx=0
Sol :
$2 \cos \frac{5 x+x}{2} \cos \frac{5 x-x}{2}+\cos 3 x=0$

2cos3xcos2x+cos3x=0

cos3x(2cos2x+1)=0

CASE-I

cos3x=0

$3 x=(2 n+1) \frac{\pi}{2} n \in z$

$x=(2 n+1) \frac{\pi}{6}, n\in z$


CASE-II
2cos2x+1=0

2cos2x=-1

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

$\cos 2 x=-\cos \frac{\pi}{3}$

$2 x=2 n \pi \pm\left(\frac{\pi}{3}\right), n \in z$

$x=n \pi \pm \frac{\pi}{6} \quad, n\in z$

Question 14

(i) sinx+sin2x+sin3x=cosx+cos2x+cos3x
Sol :
$2 \sin \frac{3 x+x}{2} \cos \frac{3 x-x}{2}+\sin 2 x=2 \cos \frac{3 x+x}{2} \cos \frac{3 x-x}{2}+\cos 2 x$

2sin2xcosx+sin2x=2cos2xcosx+cos2x

sin2x(2cosx+1)=cos2x(2cosx+1)

tan2x=1

$\tan2x=\tan\frac{\pi}{4}$

$2 x=n \pi+\frac{\pi}{4}, n \in z$

$x=\frac{n \pi}{2}+\frac{\pi}{8}, n\in z$

$=\frac{4 n \pi+\pi}{8}, n \in z$

$x=(4 n+1) \frac{\pi}{8}, n \in z$


(ii) 2sinx.sin3x=1
Sol :
2sinx.sin3x=1

cos(x-3x)-cos(x+3x)=1

cos2x-cos4x=1

cos2x=1+cos4x

cos2x=1+cos2(2x)

cos2x=2xos²2x

0=2cos²2x-cos2x

0=(2cos2x-1)cos2x

CASE-I

cos2x=0

$2 x=(2 n+1) \frac{\pi}{2}, n\in z$

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


CASE-II

2cos2x-1=0

2cos2x=1

$\cos 2 x=\frac{1}{2}$

$\cos 2 x=\cos \frac{\pi}{3}$

$2 x=2 n \pi \pm \frac{\pi}{3}, n \in z$

$x=n \pi \pm \frac{\pi}{8} , n\in z$

Question 15

(i) cos9x.cos7x=cos5x.cos3x
Sol :
2cos9x.cos7x=2cos5x.cos3x

cos(9x+7x)+cos(9x-7x)=cos(5x+3c)-cos(5x-3x)

cos16x+cos2x=cos8x+cos2x

cos16x-cos8x=0

$-2 \sin \frac{16 x+8 x}{2} \sin \frac{16 x-8 x}{2}=0$

sin12x.sin4x=0

CASE-I
sin12x=0

12x=n𝜋 , n𝜀z

$x=\frac{n \pi}{12}, n \in z$


CASE-II

sin4x=0

4x=n𝜋 , n𝜀z

$x=\frac{n \pi}{4}, n\in z$


(ii) sin5x.cos3x=sin9x.cos7x
Sol :