Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

For $x\in [0,2\pi ]$, we get

$x=\frac{\pi }{3},\frac{5\pi }{3}$

Hence, the number of values of $x$ is $2$.