Write the number of values of $x$ in $[0,2 \pi]$ that satisfy the equation $\sin x-\cos x=\frac{1}{4}$.
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.
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