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प्रश्न
Solve the following equation:
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उत्तर
Now,
\[3 ( \cos^2 x - \sin^2 x) - \sqrt{3} \sin2x = 0\]
\[ \Rightarrow 3 \cos2x - \sqrt{3} \sin2x = 0\]
\[ \Rightarrow \sqrt{3} (\sqrt{3} \cos2x - \sin2x) = 0\]
\[ \Rightarrow (\sqrt{3} \cos2x - \sin2x) = 0\]
\[ \Rightarrow \frac{\sin2x}{\cos2x} = \sqrt{3} \]
\[ \Rightarrow \tan2x = \tan \frac{\pi}{3}\]
\[ \Rightarrow 2x = n\pi + \frac{\pi}{3}, n \in Z\]
\[ \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{6}, n \in Z\]
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