Question

$\cos 3x{\cos }^{3}x+\sin 3x{\sin }^{3}x=0$.

Answer 1

$\cos 3x{\cos }^{3}x+\sin 3x{\sin }^{3}x=0$

$(4{\cos }^{3}x-3\cos x)co{s}^{3}x+(3\sin x-4{\sin }^{3}x){\sin }^{3}x=0$

$4{\cos }^{6}x-3{\cos }^{4}x+3{\sin }^{4}x-4{\sin }^{6}x=0$

$3({\sin }^{4}x-{\cos }^{4}x)-4({\sin }^{6}x-{\cos }^{6}x)=0$

$3\left[\right({\sin }^{2}x+{\cos }^{2}x\left)\right({\sin }^{2}x-{\cos }^{2}x\left)\right]\left[\right({\sin }^{2}x-{\cos }^{2}x\left)\right({\sin }^{4}x+{\cos }^{4}x+{\sin }^{2}x{\cos }^{2}x\left)\right]=0$

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

$\left(\cos 2x\right)\{-3+4[{\sin }^{2}x({\sin }^{2}x+{\cos }^{2}x)+{\cos }^{4}x\left]\right\}=0$

$\left(\cos 2x\right)\{-3+4[{\sin }^{2}x+{\cos }^{4}x\left]\right\}=0$

$\left(\cos 2x\right)\{-3+4[1-{\cos }^{2}x+{\cos }^{4}x\left]\right\}=0$

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

$\cos 2x(2{\cos }^{2}x-1{)}^2=0$

$\cos 2x\left({\cos }^{2}2x\right)=0$

${\cos }^{3}2x=0$

$\Rightarrow \cos 2x=0$

$\Rightarrow 2x=\frac{\pi }{2}+2n\pi$

$\Rightarrow x=\frac{\pi +4\pi n}{4}$