Question

The general values of $\theta$ satisfying the equation $2{\sin }^2\theta -3\sin \theta -2=0$ is

• A. $n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{\mathrm{\pi }}{6}$
• B. $n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{\mathrm{\pi }}{2}$
• C. $n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{5\mathrm{\pi }}{6}$
• D. $n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6}$

Answer 1

The correct option is D

$n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6}$

Explanation for the correct option:

Compute the general value of $\theta$.

A quadratic equation $2{\sin }^2\theta -3\sin \theta -2=0$ is given.

Find the roots of the quadratic equation.

$$\begin{gathered} \sin \theta =\frac{3\pm \sqrt{3^2-4\left(2\right)(-2)}}{2\left(2\right)} \\ \Rightarrow \sin \theta =\frac{3\pm \sqrt{9+16}}{4} \\ \Rightarrow \sin \theta =\frac{3\pm \sqrt{25}}{4} \\ \Rightarrow \sin \theta =\frac{3\pm 5}{4} \\ \Rightarrow \sin \theta =\left\{-\frac{1}{2},2\right\} \end{gathered}$$

So, the solution of the given quadratic equation is $\left\{-\frac{1}{2},2\right\}$.

Since, $\sin \theta$ can not be equal to $2$. therefore, the only solution for the quadratic equation is $\sin \theta =-\frac{1}{2}$.

$$\begin{gathered} \Rightarrow \sin \theta =\sin \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow \sin \theta =\sin \left(\frac{7\mathrm{\pi }}{6}\right) \end{gathered}$$

Therefore, the general value of $\theta$ can be given by $n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6},n\in I$.

Hence, option $D$ is the correct answer.