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Question
If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.
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Solution
The given equation is \[2 \sin^2 x = 3\cos x\].
Now,
\[2 \sin^2 x = 3\cos x\]
\[ \Rightarrow 2\left( 1 - \cos^2 x \right) = 3\cos x\]
\[ \Rightarrow 2 \cos^2 x + 3\cos x - 2 = 0\]
\[ \Rightarrow \left( 2\cos x - 1 \right)\left( \cos x + 2 \right) = 0\]
But,
cos x = -2 is not possible
\[\therefore \cos x = \frac{1}{2} = \cos\frac{\pi}{3}\]
\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in Z \left( \cos x = \cos\alpha \Rightarrow x = 2n\pi \pm \alpha, n \in Z \right)\]
Putting n = 0 and n = 1, we get
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