Question

Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]

Answer 1

\[2 \sin^2 x = 3\cos x\]
\[ \Rightarrow 2\left( 1 - \cos^2 x \right) = 3\cos x\]
\[ \Rightarrow 2 \cos^2 x + 3\cos x - 2 = 0\]
\[ \Rightarrow \left( 2\cos x - 1 \right)\left( \cos x + 2 \right) = 0\]
\[\Rightarrow \cos x = \frac{1}{2} \text{ or }\cos x = - 2\]
But,
\[\cos x = - 2\]  is not possible.

\[\left( - 1 \leq \cos x \leq 1 \right)\]
`therefore cosx=1/2=cos  pi/3`

`=>x=2npi+-pi/3,nin Z`
Putting n = 0 and n = 1, we get

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