Question

Find the general solution of the equation $2\left(\sin x-\cos 2x\right)-\sin 2x\left(1+2\sin x\right)+2\cos x=0$

Answer 1

$2(sinx-\cos 2x)-\sin 2x(1+2sinx)+2cosx=0$

$\Rightarrow 2(sinx-\cos 2x)-\sin 2x(1+2sinx)+2cosx=0$

$\Rightarrow sinx-\cos 2x-sinxcosx-{\sin }^{2}xcosx+cosx=0$

$\Rightarrow sinx-(1-{\sin }^{2}x)-sinxcosx-{\sin }^{2}xcosx+cosx=0$

$\Rightarrow sinx-1+2{\sin }^{2}x-sinxcosx-{\sin }^{2}xcosx+cosx=0$

$\{\because \cos 2x=1-2{\sin }^{2}x\&\sin 2x=2sinxcosx\}$

$\Rightarrow sinx-sinxcosx+2{\sin }^{2}x-{\sin }^{2}xcosx+cosx-1=0$

$\Rightarrow sinx(1-cosx)+2{\sin }^{2}x(1-cosx)-(1-cosx)=0$

$\Rightarrow (1-cosx)[{2\sin }^{2}x+sinx-1]=0$

$\Rightarrow 1-cosx=0$ or $2{\sin }^{2}x+sinx-1=0$

now,

${2\sin }^{2}x+sinx-1=0$

$sinx=\frac{[-1+\sqrt{1-4\left(2\right)(-1)}]}{2\times 2}$

$sinx=-1$ and $1/2$

general solution for

$cosx=1\Rightarrow x=2n\pi ,n\in z\to \left(1\right)$

$sinx=-1$ and $\Rightarrow x=m\pi +{(-1)}^m\pi /2,n\in z\to \left(2\right)$

$sinx=1/2\Rightarrow x=p\pi +{(-1)}^2\pi /6,p\in I\to \left(3\right)$

$\left(1\right)\cap \left(2\right)\cap \left(3\right)$

will be answer

general solution $=\left(2n\pi \right)\cap (m\pi +{(-1)}^m\pi /2)\cap (p\pi +{(-1)}^2\pi /6)$