Question

$$\begin{array}{cc}\text{Trigonometric Equations}& \text{General Solutions}\\ 1.\tan x=2& P.2n\pi \pm \frac{2\pi }{3},n\in I\\ 2.\sin x=\frac{\sqrt{3}}{2}& Q.n\pi -\frac{\pi }{4},n\in I\\ 3.\cos x=\frac{-1}{2}& R.n\pi +(-1{)}^n\frac{\pi }{3},n\in I\\ 4.\cot x=-1& S.n\pi +{\tan }^{-1}\left(2\right),n\in I\end{array}$$

• A. 1 - S, 2 - R, 3 - P, 4 - Q
• B. 1 - R, 2 - S, 3 - P, 4 - Q
• C. 1 - S, 2 - R, 3 - Q, 4 - P
• D. 1 - R, 2 - S, 3 - Q, 4 - P

Answer 1

The correct option is A

1 - S, 2 - R, 3 - P, 4 - Q

$1.\tan x=2$

$\tan x=\tan {\tan }^{-1}\left(2\right)$

General solution when $\tan x=\tan \alpha \text{is}x=n\pi +\alpha$

So general solution is $x=n\pi +{\tan }^{-1}\left(2\right)$ $n\in I$

$2.\sin x=\frac{\sqrt{3}}{2}$

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

So general solution is $x=n\pi +(-1{)}^n\frac{\pi }{3}$ $n\in I$

$3.\cos x=-\frac{1}{2}$

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

So general solution is $x=2n\pi \pm \frac{2\pi }{3}$ $n\in I$

$4.\cot x=-1$

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

So general solution is $x=n\pi -\frac{\pi }{4}$ $n\in I$
Hence the correct answer is Option d.