Answer 1

2
\[\sin^2 x - \cos x = \frac{1}{4}\]
\[ \Rightarrow (1 - \cos^2 x) - \cos x = \frac{1}{4}\]
\[ \Rightarrow 4 - 4 \cos^2 x - 4 \cos x = 1\]
\[ \Rightarrow 4 \cos^2 x + 4 \cos x - 3 = 0\]
\[ \Rightarrow 4 \cos^2 x + 6 \cos x - 2 \cos x - 3 = 0\]
\[ \Rightarrow 2 \cos x ( 2 \cos x + 3) - 1 ( 2 \cos x + 3) = 0\]
\[ \Rightarrow (2 \cos x + 3 ) (2 \cos x - 1) = 0\]
\[\Rightarrow 2 \cos x + 3 = 0\] or, \[2 \cos x - 1 = 0\]
\[\Rightarrow \cos x = - \frac{3}{2}\] or \[\cos x = \frac{1}{2}\]
Here,
\[\cos x = - \frac{3}{2}\]  is not possible.
\[\cos x = \frac{1}{2}\]
\[\Rightarrow \cos x = \cos \frac{\pi}{3}\]
\[ \Rightarrow x = 2n\pi \pm \frac{\pi}{3}\]
Now for n = 0 and 1, the values of \[x are \frac{\pi}{3}, \frac{5\pi}{3}\text{ and }\frac{7\pi}{3},\text{ but }\frac{7\pi}{3} \text{ is not in }\] \[\left[ 0, 2\pi \right]\]
Hence, there are two solutions in \[\left[ 0, 2\pi \right]\]