Exercise 3.8-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
Solve: $2 \cos ^2 \theta+3 \sin \theta=0$
Solution:
$
\begin{aligned}
& 2 \cos ^2 \theta+3 \sin \theta=0 \\
& \Rightarrow 2\left(1-\sin ^2 \theta\right)+3 \sin \theta=0 \\
& \Rightarrow 2-2 \sin ^2 \theta+3 \sin \theta=0 \\
& \Rightarrow-2 \sin ^2 \theta+3 \sin \theta+2=0 \\
& \Rightarrow 2 \sin ^2 \theta-3 \sin \theta-2=0
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow(2 \sin \theta+1)(\sin \theta-2)=0 \\
& \Rightarrow \quad \sin \theta=-\frac{1}{2} \quad[\because \sin \theta=2 \text { is not possible }] \\
& \Rightarrow \quad \sin \theta=-\sin \frac{\pi}{\epsilon} \\
& \Rightarrow \quad \sin \theta=\sin \left(-\frac{\pi}{6}\right) \\
& \Rightarrow \quad \theta=-\frac{\pi}{6} \\
& \Rightarrow \quad \theta=n \pi+(-1)^n \cdot\left(-\frac{\pi}{6}\right) ; n \in \mathbb{Z}
\end{aligned}
$

Question 2.
Solve: $2 \tan \theta-\cot \theta=-1$
Solution:

$
\begin{aligned}
& 2 \tan \theta-\cot \theta=-1 \\
& 2 \tan \theta-\frac{1}{\tan \theta}=-1 \Rightarrow 2 \tan ^2 \theta-1=-\tan \theta \\
& 2 \tan ^2 \theta-1+\tan \theta=0 \\
& \Rightarrow \quad(2 \tan \theta-1)(\tan \theta+1)=0 \\
& 2 \tan \theta-1=0 \text { (or) } \tan \theta+1=0 \\
& \tan \theta=\frac{1}{2} \text { (or) } \tan \theta=-1 \\
& \text { When } \tan \theta=-1=-\tan \frac{\pi}{4} \\
& \tan \theta=\tan \left(-\frac{\pi}{4}\right) \\
& \Rightarrow \theta=n \pi+\left(-\frac{\pi}{4}\right) \\
& =n \pi-\frac{\pi}{4} ; n \in \mathbb{Z} \\
& \text { When } \tan \theta=\frac{1}{2}=\tan \beta \text { (say) } \\
& \therefore \theta=n \pi+\beta=n \pi+\tan ^{-1}\left(\frac{1}{2}\right) \\
&
\end{aligned}
$
Hence $\theta=n \pi-\frac{\pi}{4}$ (or) $\theta=n \pi+\tan ^{-1}\left(\frac{1}{2}\right) ; n \in \mathbb{Z}$
Question 3.
Solve: $\tan ^2 \theta+(1-\sqrt{3})=0$
Solution:

$\begin{aligned}
& \tan ^2 \theta+\tan \theta-\sqrt{3} \tan \theta-\sqrt{3}=0 \\
& \Rightarrow \tan \theta(\tan \theta+1)-\sqrt{3}(\tan \theta+1)=0 \\
& \Rightarrow \quad(\tan \theta+1)(\tan \theta-\sqrt{3})=0
\end{aligned}$

Question 4.
Solve: $\sqrt{3} \mathrm{x}+\cos \mathrm{x}=2$
Solution:
This is of the form $a \cos x+b \sin x=c$, where $c^2 \leq a^2+b^2$
So dividing the equation by $\sqrt{(\sqrt{3})^2+1^2}$ (or) 2
we get $\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=1 \Rightarrow \sin \frac{\pi}{3} \cdot \sin x+\cos \frac{\pi}{3} \cdot \cos x=1$
$[\sin \mathrm{A} \sin \mathrm{B}+\cos \mathrm{A} \cos \mathrm{B}=\cos (\mathrm{A}-\mathrm{B})$ form]
(i.e.)
$
\begin{aligned}
\cos \left(x-\frac{\pi}{3}\right) & =1 \\
\cos \left(x-\frac{\pi}{3}\right) & =\cos 0 \\
x-\frac{\pi}{3} & =2 n \pi \pm 0 \\
x & =2 n \pi+\frac{\pi}{3}, n \in \mathbb{Z}
\end{aligned}
$

Question 5.
Solve: $\sin ^2 \theta-2 \cos \theta+\frac{1}{4}=0$
Solution:
$
\begin{aligned}
\sin ^2 \theta-2 \cos \theta+\frac{1}{4} & =0 \\
\left(1-\cos ^2 \theta\right)-2 \cos \theta+\frac{1}{4} & =0 \\
4-4 \cos ^2 \theta-8 \cos \theta+1 & =0 \\
4 \cos ^2 \theta+8 \cos \theta-5 & =0 \\
4 \cos ^2 \theta+10 \cos \theta-2 \cos \theta-5 & =0 \\
2 \cos \theta(2 \cos \theta+5)-1(2 \cos \theta+5) & =0 \\
(2 \cos \theta+5)(2 \cos \theta-1) & =0 \\
\cos \theta & =-\frac{5}{2} ; \cos \theta=\frac{1}{2}
\end{aligned}
$

$\begin{aligned}
\cos \theta & =-\frac{5}{2} \\
\therefore \cos \theta & =\frac{1}{2}=\cos \frac{\pi}{3} \\
\therefore \theta & =2 n \pi \pm \frac{\pi}{3}, n \in \mathbb{Z}
\end{aligned}$