Question

The real roots of the equation ${\cos }^{7}x+{\sin }^{4}x=1$, in the interval $(-\pi ,\pi )$ are-

• A. $0,\frac{\pi }{3},-\frac{\pi }{3}$
• B. $0,\frac{\pi }{2},-\frac{\pi }{2}$
• C. $0,\frac{\pi }{4},-\frac{\pi }{4}$
• D. None of these

Answer 1

Answer: (B)

$0,\frac{\pi }{2},-\frac{\pi }{2}$

${\cos }^{7}x+{\sin }^{4}x=1$
${\cos }^{7}x=1-{\sin }^{4}x=(1-{\sin }^{2}x)(1+{\sin }^{2}x)$

${\cos }^{7}x={\cos }^{2}x(1+{\sin }^{2}x)$
${\cos }^7\left(x\right)-{\cos }^{2}x(2-{\cos }^{2}x)=0$

${\cos }^{2}x({\cos }^{5}x+{\cos }^2(x)-2)=0$

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

$x=\frac{\pm \pi }{2}$

And

${\cos }^{5}x+{\cos }^{2}x-2=0$ implies

$\cos \left(x\right)=1$
$x=0$.

Hence

$xϵ\{\frac{-\pi }{2},0,\frac{\pi }{2}\}$