Question

Solve the equation
$sinx-3sin2x+sin3x=cosx-3cos2x+cos3x$.

• A. $\frac{n\pi }{2}-\frac{\pi }{8},n\in I$
• B. $\frac{n\pi }{2},n\in I$
• C. $\frac{n\pi }{2}+\frac{\pi }{8},n\in I$
• D. $n\pi ,n\in I$

Answer 1

Answer: (C)

Given $sinx-3sin2x+sin3x=cosx$
$-3cos2x+cos3x$
$\Rightarrow \left(sin3x+sinx\right)-3sin2x=\left(cos3x+cosx\right)-3cos2x$
$\Rightarrow 2sin2xcosx-3sin2x=2cos2xcosx-3cos2x$
$\Rightarrow sin2x\left(2cosx-3\right)=cos2x\left(2cosx-3\right)$
$\Rightarrow sin2x=cos2x\left(\because cosx\ne \frac{3}{2}\right)$
$\Rightarrow tan2x=1$
$\Rightarrow tan2x=tan\frac{\pi }{4}$
$\Rightarrow 2x=n\pi +\frac{\pi }{4}$
$\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8},n\in I$.