Question

The sum of the real roots of ${\cos }^{6}x+\sin {}^{6}x+\sin {}^{4}x=1$ in the interval$-\pi \le x\le \pi$ is equal to

• A. 0
• B. $\pi$
• C. $-\pi$
• D. none of these

Answer 1

The correct option is A 0

${\cos }^{6}x+{\sin }^{6}x=1-3{\cos }^{4}x\times {\sin }^{2}x-3{\sin }^{4}x\times {\cos }^{2}x$
$\because a^3+b^3=(a+b{)}^3-3a^{2}b-3a{b}^2$
so
$1-3{\cos }^{4}x\times {\sin }^{2}x-3{\sin }^{4}x\times {\cos }^{2}x+{\sin }^{4}x=1$
solve the above equation and get
$\sin x=0$, $\tan x=\pm \sqrt{3}$
So the some of all roots is equals to zero.