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 $
$\Rightarrow cos^7x (1 - cos^2x)^2 = 1$
$\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^2x- 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 $\cos^3x + \cos^2x -2$ = 0,
then $x$ = 0 is the only possible case.
Hence real roots are $ - \frac{\pi}{1} , 0 , \frac{\pi}{2}$