Question

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

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

Answer 1

The correct option is D $n\pi -\frac{\pi }{4}$

Given, $\sin 2x+2\sin x+2\cos x+1=0$

$\Rightarrow 1+\sin 2x+2(\sin x+\cos x)=0$

$\Rightarrow {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x+2(\sin x+\cos x)=0$

$\Rightarrow (\sin x+\cos x{)}^2+2(\sin x+\cos x)=0$

$\Rightarrow (\sin x+\cos x)(\sin x+\cos x+2)=0$

$\therefore \sin x+\cos x=0$ or $\sin x+\cos x=-2$

But, $\sin x+\cos x=-2$ is inadmissible.

Since, $|\sin x|\le 1,|\cos x|\le 1$

Therefore, $\sin x+\cos x=0$

$\Rightarrow \sin (x+\frac{\pi }{4})=0$

$\Rightarrow x+\frac{\pi }{4}=n\pi$

$\Rightarrow x=n\pi -\frac{\pi }{4}$