Question

Solve ${\cos }^{7}x+{\sin }^{4}x=1$ in the interval $(-\pi ,\pi )$

• A. $x=\frac{\pi }{2}$
• B. $x=\frac{\pi }{3}$
• C. $x=-\frac{\pi }{2}$
• D. $x=0$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $x=\frac{\pi }{2}$
C $x=-\frac{\pi }{2}$
D $x=0$

Given, ${\cos }^{7}x+{\sin }^{4}x=1$

$\Rightarrow {\cos }^{7}x+{(1-{\cos }^{2}x)}^2=1$ ....... $[sin²\theta +cos²\theta =1]$

$\Rightarrow {\cos }^{7}x+1+{\cos }^{4}x-2{\cos }^{2}x=1$

$\Rightarrow {\cos }^{7}x+{\cos }^{4}x-2{\cos }^{2}x=0$

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

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

$\Rightarrow x=\pm \frac{\pi }{2}$ or $x=0$