Question

Number of roots of ${\cos }^{7}x+{\sin }^{4}x=1$ in the interval $[-pi,+pi]$

Answer 1

${\cos }^{7}x=1-{\sin }^{4}x$

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

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

${\cos }^{2}x({\cos }^{5}x-{\sin }^{2}x-1)=0$

${\cos }^{2}x({\cos }^{5}x-1+{\cos }^{2}x-1)=0$

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

$\underset{cosx=0}{\underset{\downarrow }{{\cos }^{2}x=0}}$ or $\underset{\downarrow }{{\cos }^{5}x+{\cos }^{2}x-2=0}$

$\underset{\downarrow }{{\cos }^{2}x=1}$

$\cos x=1$ or $\cos x=-1$

Given interval is $(-\pi ,\pi )$

$\therefore x=\{-\frac{\pi }{2},\frac{\pi }{2},0,\pi ,-\pi \}$