Question

Solve the equation : $\cos 3x.{\cos }^{3}x+\sin 3x.{\sin }^{3}x=0$.

• A. $(2n+1)\frac{\pi }{4},n\in I$
• B. $(2n+1)\frac{\pi }{2},n\in I$
• C. $(2n+1)\frac{\pi }{3},n\in I$
• D. $(2n+1)\frac{\pi }{6},n\in I$

Answer 1

The correct option is A $(2n+1)\frac{\pi }{4},n\in I$

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

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

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

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

$\Rightarrow 4\left(({\cos }^{2}x-{\sin }^{2}x)({\cos }^{4}x-{\sin }^{4}x+25{\sin }^{2}x{\cos }^{2}x)\right)-3\left(({\cos }^{2}x-{\sin }^{2}x)({\cos }^{2}x+{\sin }^{2}x)\right)=0$

$($Using $a^3-b^3=(a-b)(a^2+b^2+ab)$ and $a^2-b^2=(a+b)(a-b))$

$\Rightarrow 4({\cos }^{2}x-{\sin }^{2}x)(({\cos }^{2}x+{\sin }^{2}x)-25{\sin }^{2}x{\cos }^{2}x+25{\sin }^{2}x{\cos }^{2}x)-3({\cos }^{2}x-{\sin }^{2}x)=0$

$($Using $a^2+b^2=(a+b)-2ab)$

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

$\Rightarrow {\cos }^{2}x-{\sin }^{2}x=0$

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

$\Rightarrow 2{\cos }^{2}x-1=0\Rightarrow {\cos }^{2}x=\frac{1}{2}$

$\Rightarrow x=(2n+1)\frac{\pi }{4}$