Question

The number of values of ​x in [0, 2π] that satisfy the equation ${\sin }^{2}x-\cos x=\frac{1}{4}$
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

(b) 2
$$\begin{gathered} {\sin }^{2}x-\cos x=\frac{1}{4} \\ \Rightarrow (1-{\cos }^{2}x)-\cos x=\frac{1}{4} \\ \Rightarrow 4-4{\cos }^{2}x-4\cos x=1 \\ \Rightarrow 4{\cos }^{2}x+4\cos x-3=0 \\ \Rightarrow 4{\cos }^{2}x+6\cos x-2\cos x-3=0 \\ \Rightarrow 2\cos x(2\cos x+3)-1(2\cos x+3)=0 \\ \Rightarrow (2\cos x+3)(2\cos x-1)=0 \end{gathered}$$
$\Rightarrow 2\cos x+3=0$ 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.
∴ $\cos x=\frac{1}{2}$
$$\begin{gathered} \Rightarrow \cos x=\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow x=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3} \end{gathered}$$
Now for n = 0 and 1, the values of $x\mathrm{are}\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\mathrm{and}\frac{7\mathrm{\pi }}{3},\mathrm{but}\frac{7\mathrm{\pi }}{3}\mathrm{is}\mathrm{not}\mathrm{in}$ $\left[0,2\mathrm{\pi }\right]$.
Hence, there are two solutions in $\left[0,2\mathrm{\pi }\right]$.