Solve the Following Equation: 3 Cos 2 X − 2 √ 3 Sin X Cos X − 3 Sin 2 X = 0

Solve the following equation:

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]

योग

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]

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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अध्याय 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २२]

अध्याय 11 Trigonometric equations
Exercise 11.1 | Q 3.6 | पृष्ठ २२

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