Question

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

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

Answer 1

Answer: (B)

$n\pi +{\left(-1\right)}^n\frac{\pi }{2}$

$$\begin{array}{c}A)2{\sin }^2\theta -3\sin \theta -\alpha =0\\ \Rightarrow \sin \theta =\frac{3\pm \sqrt{9+16}}{4}\\ \sin \theta =2,-\frac{1}{2}\end{array}$$

$$\begin{array}{c}\Rightarrow \sin \theta =-1/2\\ \Rightarrow \theta =n\pi +(-1{)}^n\frac{\pi }{6}\end{array}$$