Q46 / 125 · BITSAT 2016 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'D'.

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