Question

Find the general solution of: $2{\cos }^{2}x+3\sin x=0$.

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

Answer 1

Answer: (B)

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

Given:

$2{\cos }^{2}x+3\sin x=0$

Use the formula: ${\cos }^{2}x=(1-{\sin }^{2}x)$

$2(1-{\sin }^{2}x)+3\sin x=0$

$2-2{\sin }^{2}x+3\sin x=0$

$2{\sin }^{2}x-3\sin x-2=0$

$2{\sin }^{2}x-4\sin x+\sin x-2=0$

$2\sin x(\sin x-2)+1(\sin x-2)=0$

$(\sin x-2)(2\sin x+1)=0$

$\sin x-2=0or2\sin x+1=0$

$\sin x=2or\sin x=-1/2$

The value of $\sin x$ will never be 2.

$\sin x=-1/2=\sin (-\frac{\pi }{6})$

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

$x=n\pi +{(-1)}^{n+1}\left(\frac{\pi }{6}\right)$

Thus, the value of x is $n\pi -{(-1)}^n\left(\frac{\pi }{6}\right)$.