Question

If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.

Answer 1

The given equation is \[2 \sin^2 x = 3\cos x\].
Now,

\[2 \sin^2 x = 3\cos x\]

\[ \Rightarrow 2\left( 1 - \cos^2 x \right) = 3\cos x\]

\[ \Rightarrow 2 \cos^2 x + 3\cos x - 2 = 0\]

\[ \Rightarrow \left( 2\cos x - 1 \right)\left( \cos x + 2 \right) = 0\]

\[\Rightarrow \cos x = \frac{1}{2} or \cos x = - 2\]

But,
cos x = -2 is not possible

\[\left( - 1 \leq \cos x \leq 1 \right)\]

\[\therefore \cos x = \frac{1}{2} = \cos\frac{\pi}{3}\]

\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in Z \left( \cos x = \cos\alpha \Rightarrow x = 2n\pi \pm \alpha, n \in Z \right)\]
Putting n = 0 and n = 1, we get

\[x = \frac{\pi}{3}, \frac{5\pi}{3} \left( 0 \leq x \leq 2\pi \right)\]