Question

The general values of $\theta$ satisfying the equation $2si{n}^2\theta -3sin\theta -2=0$ is $(n\in Z)$

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

Answer 1

The correct option is D $n\pi +(-1{)}^n\frac{7\pi }{6}$

The given equation is
$2si{n}^2\theta -3sin\theta -2=0$
or $(2sin\theta +1)(sin\theta -2)=0$
or $sin=-\frac{1}{2}[\because sin\theta -2=0$ is not possible]
or $sin\theta =sin(-\frac{\pi }{6})=sin\left(\frac{7\pi }{6}\right)$
$\Rightarrow \theta =n\pi +(-1{)}^b(-\frac{\pi }{6})or\theta n\pi +[(-1{)}^n\frac{7\pi }{6}]$
Thus, $\theta =n\pi +(-1{)}^n\frac{7\pi }{6},n\in Z$