Exercise 3.8 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$\mathbf{E} 3.8$
Question 1.
Find the principal solution and general solutions of the following:
(i) $\sin \theta=-\frac{1}{\sqrt{2}}$
(ii) $\cot \theta=\sqrt{3}$
(iii) $\tan \theta=-\frac{1}{\sqrt{3}}$
Solution:
$(i)$ $\sin \theta=-\frac{1}{\sqrt{2}}<0$, the principal value lies in IV quadrant

$\text { Hint: } \begin{aligned}
\sin \theta & =\sin \alpha \\
\theta & =n \pi+(-1)^n \alpha
\end{aligned}$

$
\begin{aligned}
\sin \theta & =\frac{1}{\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4} \\
\text { Here } \sin \theta & =-\frac{1}{\sqrt{2}}=\sin \left(-\frac{\pi}{4}\right) \\
\Rightarrow \alpha & =-\frac{\pi}{4}
\end{aligned}
$
The general soln is $\theta=n \pi+(-1)^n \alpha$
(i.e.)
$
\theta=n \pi+(-1)^n-\frac{\pi}{4}, n \in \mathbb{Z}
$
(ii)
$
\begin{array}{rlrl}
\Rightarrow & \cot \theta & =\sqrt{3} \\
\Rightarrow & \tan \theta & =\frac{1}{\sqrt{3}}=\tan \pi / 6 \\
& & \theta & =\pi / 6
\end{array}
$
The general soln is $\theta=n \pi+\pi / 6, n \in \mathbb{Z}$
(iii)
$
\begin{aligned}
\tan \theta & =-\frac{1}{\sqrt{3}} \\
\text { Here } \tan \theta & =-\frac{1}{\sqrt{3}} \Rightarrow \theta=-\pi / 6
\end{aligned}
$
The general soln is $\theta=n \pi+(-\pi / 6)$
$
=n \pi-\pi / 6, n \in \mathbb{Z}
$

Question 2.
Solve the following equations for which solutions lies in the interval $0^{\circ}<\theta<360^{\circ}$
(i) $\sin ^4=\sin ^2 x$
(ii) $2 \cos ^2 x+1=-3 \cos x$
(iii) $2 \sin ^2 x+1=3 \sin x$
(iv) $\cos 2 x=1-3 \sin x-3 \sin x$
Solution:

$\text { (i) } \begin{aligned}
\sin ^2 x-\sin ^4 x & =0 \\
\sin ^2 x\left(1-\sin ^2 x\right) & =0 \\
\sin ^2 x\left(\cos ^2 x\right) & =0 \\
{\left[\frac{1}{2}(2 \sin x \cos x)\right]^2 } & =0 \\
\Rightarrow \quad(\sin 2 x)^2 & =0 \\
\sin 2 x & =0=0, \pi, 2 \pi, 3 \pi, 4 \pi \\
\Rightarrow \quad x & =0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi
\end{aligned}$

(ii)
$
\begin{aligned}
2 \cos ^2 x+3 \cos x+1 & =0 \\
2 \cos ^2 x+2 \cos x+\cos x+1 & =0 \\
2 \cos x(\cos x+1)+1(\cos x+1) & =0 \\
(2 \cos x+1)(\cos x+1) & =0
\end{aligned}
$

(iii)

(iv)
$
\begin{aligned}
& \sin x=\frac{1}{2}=\sin \pi / 6 \\
& x=\pi / 6 \\
& 1-2 \sin ^2 x=1-3 \sin x \\
& 3 \sin x-2 \sin ^2 x=0 \\
& \sin x(3-2 \sin x)=0 \\
& \begin{array}{l|l}
\sin x=0 & 3-2 \sin x=0 \\
\Rightarrow x=0 & \Rightarrow \sin x=3 / 2 \text { is not possible }
\end{array} \\
& \therefore \sin x=0 \Rightarrow x=0 \text { (or) } \pi \\
&
\end{aligned}
$

Question 3.
Solve the following equations:
(i) $\sin 5 x-\sin x=\cos 3 x$
(ii) $2 \cos ^2 \theta+3 \sin \theta-3=0$
(iii) $\cos \theta+\cos 3 \theta=2 \cos 2 \theta$
(iv) $\sin \theta+\sin 3 \theta+\sin 5 \theta=0$
(v) $\sin 2 \theta-\cos 2 \theta-\sin \theta+\cos \theta=0$
(vi) $\sin \theta+\cos \theta=\sqrt{3}$
(vii) $\sin \theta+\sqrt{3} \cos \theta=1$
(viii) $\cot \theta+\operatorname{cosec} \theta=\sqrt{3}$
(ix) $\tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=\sqrt{3}$
(x) $\cos 2 \theta=\frac{\sqrt{5}+1}{4}$
(xi) $2 \cos ^2 x-7 \cos x+3=0$
Solution:

$\begin{aligned}
& \text { (i) } \sin 5 x-\sin x=\cos 3 x \\
& 2 \cos \frac{5 x+x}{2} \sin \frac{5 x-x}{2}=\cos 3 x
\end{aligned}$

The general soln is
$
x=(2 n+1) \pi / 6 \text { (or) } x=n \pi / 2+(-1)^n \pi / 12, n \in \mathbb{Z}
$
(ii) $2 \cos ^2 \theta+3 \sin \theta-3=0$
$
\begin{array}{r}
2\left(1-\sin ^2 \theta\right)+3 \sin \theta-3=0 \\
2-2 \sin ^2 \theta+3 \sin \theta-3=0
\end{array}
$

The general soln is
$
\theta=n \pi+(-1)^n \pi / 2 ; \theta=n \pi+(-1)^n \pi / 6, n \in \mathbb{Z}
$
(iii) $\cos \theta+\cos 3 \theta=2 \cos 2 \theta$
$
2 \cos \frac{\theta+3 \theta}{2} \cos \frac{3 \theta-\theta}{2}=2 \cos 2 \theta
$

(iv)

$\begin{aligned}
& \sin \theta+\sin 3 \theta+\sin 5 \theta=0 \\
&(\sin \theta+\sin 5 \theta)+\sin 3 \theta=0 \\
& 2 \sin \frac{5 \theta+\theta}{2} \cos \frac{5 \theta-\theta}{2}+\sin 3 \theta=0 \\
& 2 \sin 3 \theta \cos 2 \theta+\sin 3 \theta=0 \\
& \sin 3 \theta[2 \cos 2 \theta+1]=0
\end{aligned}$

The general soln is $\theta=n \pi / 3$ (or) $\theta=n \pi \pm \pi / 3, n \in \mathrm{Z}$
(v) $(\sin 2 \theta-\sin \theta)-(\cos 2 \theta-\cos \theta)=0$
$
\begin{aligned}
2 \cos \frac{3 \theta}{2} \sin \frac{\theta}{2}-\left(-2 \sin \frac{3 \theta}{2} \sin \frac{\theta}{2}\right) & =0 \\
2 \sin \frac{\theta}{2}\left[\cos \frac{3 \theta}{2}+\sin \frac{3 \theta}{2}\right] & =0
\end{aligned}
$

The general soln is $\theta=2 n \pi$ (or) $\frac{2 n \pi}{3}-\frac{\pi}{6}, n \in \mathrm{Z}$
(vi) $\sin \theta+\cos \theta=\sqrt{2}$
Multiplying by $\frac{1}{\sqrt{2}}$ on both sides
$
\begin{aligned}
\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta & =\frac{1}{\sqrt{2}} \sqrt{2}=1 \\
\cos \theta \cos \frac{\pi}{4}+\sin \theta \sin \frac{\pi}{4} & =1
\end{aligned}
$
The gẹneral soln is $\cos (\theta-\pi / 4)=1$
$
\begin{aligned}
\cos (\theta-\pi / 4) & =\cos 0 \\
\therefore \theta-\pi / 4 & =2 n \pi \pm 0
\end{aligned}
$
The general soln is $\theta=2 n \pi+\pi / 4, n \in \mathrm{Z}$
$
=\frac{8 n \pi+\pi}{4}=\frac{\pi}{4}(8 n+1), n \in z
$
(vii) $\sin \theta+\sqrt{3} \cos \theta=1$
Multiplying by $\frac{1}{\sqrt{1+3}}=\frac{1}{2}$ we get
$
\begin{aligned}
\frac{1}{2} \sin \theta+\frac{\sqrt{3}}{2} \cos \theta & =\frac{1}{2} \\
\cos \theta \cos \frac{\pi}{6}+\sin \theta \sin \frac{\pi}{6} & =\frac{1}{2}
\end{aligned}
$

$\begin{aligned}
& \cos (\theta-\pi / 6)=\frac{1}{2}=\cos \pi / 3 \\
& \Rightarrow \quad \theta-\pi / 6=\pi / 3 \\
& \text { General soln is } \theta-\pi / 6=2 n \pi \pm \pi / 3 \\
& \theta=2 n \pi+\pi / 6 \pm \pi / 3, n \in \mathrm{Z} \\
& \text { (viii) } \frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}=\sqrt{3} \\
& \cos \theta+1=\sqrt{3} \sin \theta \\
& \sqrt{3} \sin \theta-\cos \theta=1 \\
& (\div \text { by } 2) \frac{\sqrt{3}}{2} \sin \theta-\frac{1}{2} \cos \theta=\frac{1}{2} \\
& \text { (i.e) } \cos \theta \cos \frac{\pi}{3}-\sin \theta \sin \frac{\pi}{3}=-\frac{1}{2} \\
& \Rightarrow \quad \cos \left(\theta+\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} \\
& \Rightarrow \quad \theta+\frac{\pi}{3}=\frac{2 \pi}{3} \\
&
\end{aligned}$

The general soln is $\theta=2 n \pi \pm \frac{2 \pi}{3}-\frac{\pi}{3}, n \in \mathrm{Z}$
(ix) Now $\tan \left(\theta+\frac{\pi}{3}\right)=\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}$ and $\tan \left(\theta+\frac{2 \pi}{3}\right)=\frac{\tan \theta-\sqrt{3}}{1+\sqrt{3} \tan \theta}$
So, $\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)$
$
\begin{aligned}
& =\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}+\frac{\tan \theta-\sqrt{3}}{1+\sqrt{3} \tan \theta}=\frac{(\tan \theta+\sqrt{3})(1+\sqrt{3} \tan \theta)+(\tan \theta-\sqrt{3})(1-\sqrt{3} \tan \theta)}{1-3 \tan ^2 \theta} \\
& =\frac{\tan \theta+\sqrt{3}+\sqrt{3} \tan ^2 \theta+3 \tan \theta+\tan \theta-\sqrt{3} \tan ^2 \theta-\sqrt{3}+3 \tan \theta}{1-3 \tan ^2 \theta}=\frac{8 \tan \theta}{1-3 \tan ^2 \theta} \\
& \text { Given, } \tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=\sqrt{3} \\
& \Rightarrow \quad \tan \theta+\frac{8 \tan \theta}{1-3 \tan ^2 \theta}=\sqrt{3} \Rightarrow \frac{\tan \theta-3 \tan ^3 \theta+8 \tan \theta}{1-3 \tan ^2 \theta}=\sqrt{3}
\end{aligned}
$

$
\begin{aligned}
& \text { (x) } \cos 2 \theta=\frac{\sqrt{5}+1}{4}=\cos 36^{\circ}=\cos \frac{2 \pi}{10} \\
& \Rightarrow \quad 2 \theta=\frac{2 \pi}{10} \\
& \text { The general soln is } 2 \theta=2 n \pi \pm \frac{2 \pi}{10} \\
& \theta=n \pi \pm \frac{\pi}{10}, n \in \mathbb{Z} \\
&
\end{aligned}
$
(xi)
$
\begin{aligned}
2 \cos ^2 x-7 \cos x+3 & =0 \\
\cos x & =\frac{7 \pm \sqrt{49-24}}{2(2)}=\frac{7 \pm 5}{4} \\
\cos x & =\frac{7+5}{4}, \frac{7-5}{4} \\
\cos x & =3 \text { (or) } \frac{1}{2} \\
\cos x & =3 \text { is not possible } \\
\cos x & =\frac{1}{2}=\cos \frac{\pi}{3} \\
\Rightarrow \quad x & =\pi / 3
\end{aligned}
$
The general soln is $x=2 n \pi \pm \pi / 3, n \in \mathrm{Z}$