Question

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

Answer 1

(b) 2
$$\begin{gathered} {\sin }^2\theta -\cos \theta =\frac{1}{4} \\ \Rightarrow (1-{\cos }^2\theta )-\cos \theta =\frac{1}{4} \\ \Rightarrow 4-4{\cos }^2\theta -4\cos \theta =1 \\ \Rightarrow 4{\cos }^2\theta +4\cos \theta -3=0 \\ \Rightarrow 4{\cos }^2\theta +6\cos \theta -2\cos \theta -3=0 \\ \Rightarrow 2\cos \theta (2\cos \theta +3)-1(2\cos \theta +3)=0 \\ \Rightarrow (2\cos \theta +3)(2\cos \theta -1)=0 \end{gathered}$$
$\Rightarrow 2\cos \theta +3=0$ or, $2\cos \theta -1=0$
$\Rightarrow \cos \theta =-\frac{3}{2}$ or $\cos \theta =\frac{1}{2}$
Here, $\cos \theta =-\frac{3}{2}$ is not possible.
∴ $\cos \theta =\frac{1}{2}$

$$\begin{gathered} \Rightarrow \cos \theta =\cos \frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3} \end{gathered}$$

Now for n = 0 and 1, the values of $\theta \mathrm{are}\frac{\mathrm{\pi }}{3},\frac{5\mathrm{\pi }}{3}\mathrm{and}\frac{7\mathrm{\pi }}{3},\mathrm{but}\frac{7\mathrm{\pi }}{3}\mathrm{is}\mathrm{not}\mathrm{in}$ $\left[0,2\mathrm{\pi }\right]$.
Hence, there are two solutions in $\left[0,2\mathrm{\pi }\right]$.