Question

If ${\sin }^2\left(\theta \right)=\frac{1}{4}$, then the most general value of $\left(\theta \right)$ is

• A. $2n\mathrm{\pi }\pm {\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{6}$
• B. $\left(\frac{n\mathrm{\pi }}{2}\right)\pm {\left(-1\right)}^n\frac{\mathrm{\pi }}{6}$
• C. $n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6}$
• D. $2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6}$

Answer 1

The correct option is C

$n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6}$

Step 1: Given information

${\sin }^2\left(\theta \right)=\frac{1}{4}........\left(1\right)$

Step 2: Simplification and calculate the most general value of $\left(\theta \right)$

Take square root of equation 1

$\begin{array}{rcl}& \Rightarrow & \sin \left(\theta \right)=\sqrt{\frac{1}{4}}\\ & \Rightarrow & \sin \left(\theta \right)=\pm \frac{1}{2}\\ & & \\ & \Rightarrow & \sin \left(\theta \right)=\pm \sin \left(\frac{\mathrm{\pi }}{6}\right)\\ & & \end{array}$

For the general value $\left(\theta \right)$,

$\begin{array}{rcl}& \Rightarrow & \left(\theta \right)=n\pi \pm \frac{\mathrm{\pi }}{6}\\ & & \end{array}$

Hence , option $\left(\mathit{C}\right)$ is correct.