Question

Solve the equation
$2(cos \,x + cos2x) + sin2x( 1 + 2cosx) = 2sinx$.

• A. $n\pi+\left(-1\right)^{n}\left(-\frac{\pi}{2}\right)$, $n \in I$
• B. $2n\pi \pm\pi$ or $2n\pi\pm\frac{\pi}{3}$, $n \in I$
• C. $ n\pi$, $n\in I$
• D. Both $(a)$ and $(b)$

Answer 1

Answer: (D)

The given equation is
$2(cosx + cos2x) + sin2x(1 + 2cosx) = 2sinx$
$\Rightarrow 2\left(cosx + 2cos^{2}x - 1\right) + 2sinxcosx\left(1 + 2 \,cosx\right)$
$= 2sinx$
$\Rightarrow 2cosx + 2sinx \,cosx + 4cos^{2}x + 4sinxcos^{2}x - 2$
$-2sinx=0$
$\Rightarrow 2cosx\left(1 + sinx\right) + 4cos^{2}x\left(1 + sinx\right) -2\left(1 + sinx\right)$
$ = 0$
$\Rightarrow 2\left(1 + sinx\right)\left\{cosx + 2cos^{2}x - 1\right\} = 0$
$\Rightarrow $ either $1 + sinx = 0$ or $2cos^{2}x + cosx - 1 = 0$
First, consider $1 + sinx = 0$
$\Rightarrow sinx=-1=sin\left(-\frac{\pi}{2}\right)$
$\Rightarrow x=n\pi+\left(-1\right)^{n}\left(-\frac{\pi}{2}\right)$, $n \in I$.
Next, consider $2cos^{2}\,x+cosx-1=0$
$\Rightarrow cosx=\frac{-1 \pm\sqrt{1+8}}{4}$
$=\frac{-1 \pm3}{4}=1$, $\frac{1}{2}$
$\Rightarrow cosx=cos\pi$, $cos \frac{\pi}{3}$
$\Rightarrow x=2n\pi \pm\pi$ or $2n\pi \pm\frac{\pi}{3}$, $n \in I$.