Q63 / 160 · AP EAMCET 2021 (23 Aug Shift 2) · Mathematics

Solve the following equation $\sin x+\sqrt{3} \cos x=\sqrt{2}$

  • Correct answer
    A
    $x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{4}$
  • B
    $x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{3}$
  • C
    $x=0$
  • D
    $x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{2}$

Solution

$\sin x+\sqrt{3} \cos x=\sqrt{2}$ $\Rightarrow \quad \sin x=\sqrt{2}-\sqrt{3} \cos x$ On squaring both sides, $\sin ^2 x=2+3 \cos ^2 x-2 \sqrt{6} \cos x$ $\Rightarrow \quad 1-\cos ^2 x=2+3 \cos ^2 x-2 \sqrt{6} \cos x$ $\Rightarrow \quad 4 \cos ^2 x-2 \sqrt{6} \cos x+1=0$ ...(i) $\Rightarrow \quad \cos x=\frac{2 \sqrt{6} \pm \sqrt{24-16}}{2 \times 4}$ $\Rightarrow \quad=\frac{2 \sqrt{6} \pm 2 \sqrt{2}}{8}$ $\Rightarrow \quad \cos x=\frac{\sqrt{6} \pm \sqrt{2}}{4}=\cos \left(\frac{\pi}{6} \pm \frac{\pi}{4}\right)$ $\Rightarrow \quad x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{4}$

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

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