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# The number of values of ​x in [0, 2π]

Question:

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

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

(b) 2

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

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

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

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

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

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

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

$\Rightarrow 2 \cos x+3=0$ 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.

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

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

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

Now for $n=0$ and 1, the values of $x$ are $\frac{\pi}{3}, \frac{5 \pi}{3}$ and $\frac{7 \pi}{3}$, but $\frac{7 \pi}{3}$ is not in $[0,2 \pi]$.

Hence, there are two solutions in $[0,2 \pi]$.