Question

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]

Answer 1

\[\sin^2 x - \cos x = \frac{1}{4}\]
\[\Rightarrow 1 - \cos^2 x - \cos x = \frac{1}{4}\]
\[ \Rightarrow 4 - 4 \cos^2 x - 4 \cos x = 1\]
\[ \Rightarrow 4 \cos^2 x + 4 \cos x - 3 = 0\]
\[ \Rightarrow 4 \cos^2 x + 6 \cos x - 2 \cos x - 3 = 0\]
\[ \Rightarrow 2 \cos x(2 \cos x + 3) - 1(2 \cos x + 3) = 0\]
\[ \Rightarrow (2 \cos x + 3) (2 \cos x - 1) = 0\]

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

\[2 \cos x + 3 = 0\]

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

\[\cos x = - \frac{3}{2}\] is not possible.
\[\therefore \cos x = \frac{1}{2} \]
\[ \Rightarrow \cos x = \cos\frac{\pi}{3} \]
\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in Z\]