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