Question

If $0\le x\le 2\pi$, then the number of solutions of the equation ${\sin }^{6}x+{\cos }^{6}x=1$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$
• E. $8$

Answer 1

The correct option is B $3$

Given equation is ${\sin }^{8}x+{\cos }^{6}x=1$
$\Rightarrow {\sin }^{8}x+{(1-{\sin }^{2}x)}^3=1$
$\Rightarrow {\sin }^{8}x-{\sin }^{6}x+3{\sin }^{4}x-3{\sin }^{2}x=0$
$\Rightarrow {\sin }^{6}x({\sin }^{2}x-1)+3{\sin }^{2}x({\sin }^{2}x-1)=0$
$\Rightarrow ({\sin }^{6}x+3{\sin }^{2}x)({\sin }^{2}x-1)=0$
$\Rightarrow {\sin }^{2}x({\sin }^{4}x+3)({\sin }^{2}x-1)=0$
$\Rightarrow ({\sin }^{4}x+3)\left[{\sin }^{2}x(\sin x-1)(\sin x+1)\right]=0$
$\Rightarrow {\sin }^{2}x(\sin x-1)(\sin x+1)=0$ $[\because {\sin }^{4}x+3\ne 0]$
From above, it is clear that it has three roots in $[0,2\pi ]$.