Question

Find the general solution of $sinx+\sqrt{2}=-sinx$

• A. $x=n\pi +(-1{)}^{n+1}\frac{\pi }{4}$
• B. $x=n\pi +(-1{)}^n\frac{\pi }{4}$
• C. $x=n\pi +(-1{)}^n\frac{\pi }{3}$
• D. $x=n\pi +(-1{)}^n\frac{3\pi }{4}$

Answer 1

The correct option is A

$x=n\pi +(-1{)}^{n+1}\frac{\pi }{4}$

$sinx+\sqrt{2}=-sinx2sinx=-\sqrt{2}sinx=-\frac{1}{\sqrt{2}}x=n\pi +(-1{)}^n(-\frac{\pi }{4})x=n\pi +(-1{)}^{n+1}\frac{\pi }{4}$