Write the number of values of x in [0, 2π]

Solution:

Given equation: $\sin ^{2} x-\cos x=\frac{1}{4}$

Now,

$\left(1-\cos ^{2} x\right)-\cos x=\frac{1}{4}$

$\Rightarrow 4-4 \cos ^{2} x-4 \cos x=1$

$\Rightarrow 4 \cos ^{2} x+4 \cos x-3=0$

$\Rightarrow 4 \cos ^{2} x+6 \cos x-2 \cos x-3=0$

$\Rightarrow 2 \cos x(2 \cos x+3)-1(2 \cos x+3)=0$

$\Rightarrow(2 \cos x+3)(2 \cos x-1)=0$

Here, $2 \cos x+3=0 \Rightarrow \cos x=-\frac{3}{2}$ is not possible.

Or,

$2 \cos x-1=0$

$\Rightarrow \cos x=\frac{1}{2}$

$\Rightarrow \cos x=\cos \frac{\pi}{3}$

$\Rightarrow x=2 n \pi \pm \frac{\pi}{3}$

Taking positive sign, $x=\frac{7 \pi}{3}, \frac{13 \pi}{3}, \frac{19 \pi}{3}, \ldots$

Taking negative sign, $x=\frac{5 \pi}{3}, \frac{11 \pi}{3}, \frac{17 \pi}{3}, \ldots$

$x=\frac{5 \pi}{3}$ and $\frac{7 \pi}{3}$ will satisfy the given condition, i.e., $x$ in $[0,2 \pi]$.

Hence, two values will satisfy the given equation.