Question

Solve the equation
$2(cosx+cos2x)+sin2x(1+2cosx)=2sinx$.

• 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
$2(cosx+cos2x)+sin2x(1+2cosx)=2sinx$
$\Rightarrow 2\left(cosx+2co{s}^{2}x-1\right)+2sinxcosx\left(1+2cosx\right)$
$=2sinx$
$\Rightarrow 2cosx+2sinxcosx+4co{s}^{2}x+4sinxco{s}^{2}x-2$
$-2sinx=0$
$\Rightarrow 2cosx\left(1+sinx\right)+4co{s}^{2}x\left(1+sinx\right)-2\left(1+sinx\right)$
$=0$
$\Rightarrow 2\left(1+sinx\right)\left\{cosx+2co{s}^{2}x-1\right\}=0$
$\Rightarrow$ either $1+sinx=0$ or $2co{s}^{2}x+cosx-1=0$
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 $2co{s}^{2}x+cosx-1=0$
$\Rightarrow cosx=\frac{-1\pm \sqrt{1+8}}{4}$
$=\frac{-1\pm 3}{4}=1$, $\frac{1}{2}$
$\Rightarrow cosx=cos\pi$, $cos\frac{\pi }{3}$
$\Rightarrow x=2n\pi \pm \pi$ or $2n\pi \pm \frac{\pi }{3}$, $n\in I$.