Question

General solution of the equation ${\cot }^2\theta +\frac{3}{\sin \theta }+3=0$ is

• A. $n\pi +(-1{)}^n(-\frac{\pi }{6})$
• B. $n\pi +(-1{)}^n(-\frac{\pi }{4})$
• C. $n\pi$
• D. $Noneofthese$

Answer 1

Answer: (A)

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

Given ${\cot }^2\theta +\frac{3}{\sin \theta }+3=0$

As we know that ${\cot }^2\theta ={\text{cosec}}^2\theta -1$

${\text{cosec}}^2\theta -1+3\text{cosec}\theta +3=0$

$⟹{\text{cosec}}^2\theta +3\text{cosec}\theta +2=0$

$⟹(\text{cosec}\theta +2)(\text{cosec}\theta +1)=0$

$⟹\text{cosec}\theta =-2$ or $-1$

So $x=n\pi +(-1{)}^n(-\frac{\pi }{6})$ or $\frac{(2k+3)\pi }{2}$