Question

Solve the following equation:

\[4 \sin^2 x - 8 \cos x + 1 = 0\]

Answer 1

\[4 \sin^2 x - 8\cos x + 1 = 0\]

\[\Rightarrow 4 - 4 \cos^2 x - 8 \cos x + 1 = 0\]
\[ \Rightarrow 4 \cos^2 x + 8 \cos x - 5 = 0\]
\[ \Rightarrow 4 \cos^2 x + 10 \cos x - 2 \cos x - 5 = 0\]
\[ \Rightarrow 2 \cos x (2 \cos x + 5 ) - 1 (2 \cos x + 5) = 0\]
\[ \Rightarrow (2 \cos x - 1) (2 \cos x + 5) = 0\]

\[\Rightarrow (2 \cos x - 1) = 0\] or

\[(2 \cos x + 5) = 0\]

Now,

\[2 \cos x + 5 = 0 \Rightarrow \cos x = - \frac{5}{2}\]  (It is not possible.)

\[\therefore 2 \cos x - 1 = 0 \]

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

\[ \Rightarrow \cos x = \cos \frac{\pi}{3} \]

\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in Z\]