Question

Solve: $2(\sin x-\cos 2x)-\sin 2x(1+2\sin x)+2\cos x=0$

Answer 1

Solving $\mathbf{2}\mathbf{(}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{)}\mathbf{}\mathbf{-}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$:

$\Rightarrow 2\sin x-\sin 2x-2\cos 2x-2\sin x\sin 2x+2\cos x=0$

We know that, $\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}$

So, $2\sin x\sin 2x=2\sin x(2\sin x\cos x)$

$=4{\sin }^{2}x\cos x$

$=4(1–{\cos }^{2}x)\cos x$ $\left({\sin }^{2}x=1-{\cos }^{2}x\right)$

$=4\cos x–4{\cos }^{3}x$

$=\cos x+(3\cos x–4{\cos }^{3}x)$

$=\cos x-\cos 3x$ $(\cos 3x=-3\mathrm{cosx}+4{\cos }^{3}x)$

So, $2\sin x-2\sin x\cos x-2\cos 2x-(\cos x-\cos 3x)+2\cos x=0$

$\Rightarrow 2\sin x(1-\cos x)+4{\cos }^{3}x-3\cos x+\cos x-2(2{\cos }^{2}x-1)=0$

$\Rightarrow 2\sin x(1-\cos x)+4{\cos }^{3}x-4{\cos }^{2}x-2\cos x+2=0$

$\Rightarrow 2\sin x(1-\cos x)-4{\cos }^{2}x(1-\cos x)+2(1-\cos x)=0$

$\Rightarrow (1-\cos x)[2\sin x-4(1-{\sin }^{2}x)+2]=0$

$\text{⇒}{\text{cos x = 1 or sin x − 2(1 − sin}}^{2}x)+1=0$

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

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

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

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

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

$\Rightarrow$ $x=m\mathrm{\pi }+{(-1)}^{\mathrm{m}}\frac{\mathrm{\pi }}{6}\mathrm{or}\mathrm{x}=\mathrm{k\pi }+{(-1)}^{\mathrm{k}}\frac{-\mathrm{\pi }}{2}$

Hence general solutions are $x=2n\mathrm{\pi }orx=m\mathrm{\pi }+{(-1)}^{\mathrm{m}}\frac{\mathrm{\pi }}{6}\mathrm{or}\mathrm{x}=\mathrm{k\pi }+{(-1)}^{\mathrm{k}}\frac{-\mathrm{\pi }}{2}$