Question

The general solution of the equation $\sin \, 2x + 2 \sin \, x +2 \cos \, x + 1 = 0$ is

• A. $3 n \pi - \frac{\pi}{4}$
• B. $2 n \pi + \frac{\pi}{4}$
• C. $2 n \pi +(-1)^n \, \sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$
• D. $n \pi - \frac{\pi}{4}$

Answer 1

Answer: (D)

Given, $\sin 2 x+2 \sin x+2 \cos x+1=0$
$\Rightarrow 1+\sin 2 x+2(\sin x+\cos x)=0$
$\Rightarrow (\sin x+\cos x)^{2}+2(\sin x+\cos x)=0$
$\Rightarrow (\sin x+\cos x)(\sin x+\cos x+2)=0$
$\sin x+\cos x=0$ or $\sin x+\cos x=-2$
But, $\sin x+\cos x=-2$ is inadmissible.
Since, $|\sin x| \leq 1,|\cos x| \leq 1$
$\therefore \sin x+\cos x=0 \Rightarrow \sin \left(x+\frac{\pi}{4}\right)=0$
$\Rightarrow x+\frac{\pi}{4}=n \pi \Rightarrow x=n \pi-\frac{\pi}{4}$