Question

If ${\sin }^2\left(\theta \right)-2\cos \left(\theta \right)+\frac{1}{4}=0$, then the general value of $\left(\theta \right)$ is

• A. $n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}$
• B. $2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}$
• C. $2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6}$
• D. $n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6}$

Answer 1

The correct option is B

$2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}$

Explanation for the correct option

Step 1 : Simplification and calculate the general value of $\left(\theta \right)$

Use the identity, ${\sin }^2\left(\theta \right)=1-{\cos }^2\left(\theta \right)$

put this value in equation $\left(1\right)$ in place of ${\sin }^2\left(\theta \right)$

$\begin{array}{rcl}& \Rightarrow & 1-{\cos }^2\left(\theta \right)-2\cos \left(\theta \right)+\frac{1}{4}=0\\ & \Rightarrow & -{\cos }^2\left(\theta \right)-2\cos \left(\theta \right)+\frac{5}{4}=0\\ & \Rightarrow & {\cos }^2\left(\theta \right)+2\cos \left(\theta \right)-\frac{5}{4}=0\end{array}$

Factor of $2\cos \left(\theta \right)$,

$\begin{array}{rcl}& \Rightarrow & {\cos }^2\left(\theta \right)-\left(\frac{1}{2}\right)\cos \left(\theta \right)+\left(\frac{5}{2}\right)\cos \left(\theta \right)-\frac{5}{4}=0\\ & \Rightarrow & \cos \left(\theta \right)\left[\cos \left(\theta \right)-\left(\frac{1}{2}\right)\right]+\left(\frac{5}{2}\right)\left[\cos \left(\theta \right)-\left(\frac{1}{2}\right)\right]=0\\ & \Rightarrow & \left[\cos \left(\theta \right)+\left(\frac{5}{2}\right)\right]\left[\cos \left(\theta \right)-\left(\frac{1}{2}\right)\right]=0\\ & & \end{array}$

Take

$\begin{array}{rcl}& \Rightarrow & \cos \left(\theta \right)+\left(\frac{5}{2}\right)=0\\ & \Rightarrow & \cos \left(\theta \right)=-\left(\frac{5}{2}\right)\end{array}$

And

$\begin{array}{rcl}& \Rightarrow & \cos \left(\theta \right)-\left(\frac{1}{2}\right)=0\\ & \Rightarrow & \cos \left(\theta \right)=\left(\frac{1}{2}\right)\\ & & \end{array}$

After that ,

$\Rightarrow \cos \left(\theta \right)=-\left(\frac{5}{2}\right)$ is not possible because the range of $\cos \left(\theta \right)$ is $\left[-1,1\right]$

So , we take the value of

$\Rightarrow \cos \left(\theta \right)=\left(\frac{1}{2}\right).....\left(2\right)$

Step : Formula for the general value of $\cos \left(\theta \right)=2n\pi \pm \theta$ put in equation $\left(2\right)$, then

$\begin{array}{rcl}& \Rightarrow & \cos \left(\theta \right)=\cos \left(\frac{\mathrm{\pi }}{3}\right)\\ & \Rightarrow & \theta =2n\pi \pm \frac{\mathrm{\pi }}{3}\end{array}$

This is the general value of $\left(\theta \right)$.

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