AP EAMCET · Maths · Trigonometric Ratios & Identities
Solve the following equation \(\sin x+\sqrt{3} \cos x=\sqrt{2}\)
- 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 \(x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{2}\)
Answer & Solution
Correct Answer
(A) \(x=2 n \pi+\frac{\pi}{6} \pm \frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\sin x+\sqrt{3} \cos x=\sqrt{2}\) \(\Rightarrow \quad \sin x=\sqrt{2}-\sqrt{3} \cos x\) On squaring both sides, \(\sin ^2 x=2+3 \cos ^2 x-2 \sqrt{6} \cos x\) \(\Rightarrow \quad 1-\cos ^2 x=2+3 \cos ^2 x-2 \sqrt{6} \cos x\)…
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