Solve the following equation $\sin x+\sqrt{3} \cos x=\sqrt{2}$
Solve the following equation $\sin x+\sqrt{3} \cos x=\sqrt{2}$
- $x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{4}$
- $x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{3}$
- $x=0$
- $x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{2}$
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)
$\Rightarrow \quad \cos x=\frac{2 \sqrt{6} \pm \sqrt{24-16}}{2 \times 4}$
$\Rightarrow \quad=\frac{2 \sqrt{6} \pm 2 \sqrt{2}}{8}$
$\Rightarrow \quad \cos x=\frac{\sqrt{6} \pm \sqrt{2}}{4}=\cos \left(\frac{\pi}{6} \pm \frac{\pi}{4}\right)$
$\Rightarrow \quad x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{4}$
Asked in: AP EAMCET 2021 (23 Aug Shift 2)
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