Question

The solution set of $\sin (x+\frac{\pi }{4})=\sin 2x$ equals to

• A. $\frac{n\pi -\frac{\pi }{4}}{1+(-1{)}^{n}2}$
• B. $\frac{n\pi +\frac{\pi }{4}}{1+(-1{)}^{n}2}$
• C. $\frac{n\pi +\frac{\pi }{4}}{1-(-1{)}^{n}2}$
• D. $\frac{n\pi -\frac{\pi }{4}}{1-(-1{)}^{n}2}$

Answer 1

Answer: (D)

$\frac{n\pi -\frac{\pi }{4}}{1-(-1{)}^{n}2}$

The general solution of $sinx=sin\alpha$ is,
$x=n\pi +{(-1)}^n\alpha$
Similarly,
The solution of $sin(x+\frac{\pi }{4})=sin2x$ is,
$(x+\frac{\pi }{4})=n\pi +{(-1)}^n\left(2x\right)$
$x=\frac{n\pi -\frac{\pi }{4}}{1-{(-1)}^n\left(2\right)}$