Question

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

• A. $-\frac{\pi }{2},0$
• B. $-\frac{\pi }{2},0,\frac{\pi }{2}$
• C. $\frac{\pi }{2},0$
• D. $0,\frac{\pi }{4},\frac{\pi }{2}$

Answer 1

Answer: (B)

${\cos }^{7}x+{\sin }^{4}x=1$
$\Rightarrow co{s}^{7}x(1-co{s}^{2}x{)}^2=1$
$\Rightarrow {\cos }^{7}x+1+{\cos }^{4}x-2{\cos }^{2}x=1$
$\Rightarrow {\cos }^{7}x+1+{\cos }^{4}x-2{\cos }^{2}x=1$
$\Rightarrow {\cos }^{7}x+{\cos }^{4}x-2{\cos }^{2}x=0$
$\Rightarrow {\cos }^{2x}\left[{\cos }^{5}x+{\cos }^{2}x-2\right]=0$
If ${\cos }^{2}x$ = 0, then $x=n\pi \pm \frac{\pi }{2},n\in I$
$\therefore$ $x=\pm \frac{\pi }{2}$ $[\because -\pi <x<\pi ]$
and if ${\cos }^{3}x+{\cos }^{2}x-2$ = 0,
then $x$ = 0 is the only possible case.
Hence real roots are $-\frac{\pi }{1},0,\frac{\pi }{2}$