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. $4$

Answer 1

The correct option is B $2$

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

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

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

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

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

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

$\Rightarrow (2\cos x-1)=0\text{or}(2\cos x+3)=0$

$\Rightarrow cosx=\frac{1}{2}\text{or}\cos x=-\frac{3}{2}$

As $\cos x\in [-1,1]$, so $\cos x=\frac{1}{2}$

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

We know the general solution of $\cos x=\cos \alpha$ is $x=2n\pi \pm \alpha ,n\in Z$

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

$\therefore x=\frac{\pi }{3},\frac{5\pi }{3}$
So, the number of solution is $2$.
Hence, the option (B) is correct.