Question

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

• A. one
• B. three
• C. two
• D. four

Answer 1

The correct option is B three

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

$\Rightarrow {\cos }^{7}x-1+{\sin }^{4}x=0$

$\Rightarrow {\cos }^{7}x-(1-{\sin }^{4}x)=0$

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

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

$\Rightarrow {\cos }^{2}x[{\cos }^{5}x-1-(1-{\cos }^{2}x)]=0$

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

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

$\therefore {\cos }^{2}x=0$

$\Rightarrow x=\frac{-\pi }{2}\&\frac{\pi }{2}\in (-\pi ,\pi )$

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

$\Rightarrow {\cos }^{5}x-{\cos }^{4}x+{\cos }^{4}x-{\cos }^{3}x+{\cos }^{3}x-{\cos }^{2}x2{\cos }^{2}x-2\cos x+2\cos x-2=0$

$\Rightarrow {\cos }^{4}x(\cos x-1)+{\cos }^{3}x(\cos x-1)+{\cos }^{2}x(\cos x-1)+2\cos x(\cos x-1)+2(\cos x-1)=0$

$\Rightarrow (\cos x-1)({\cos }^{4}x+{\cos }^{3}x+{\cos }^{2}x+2\cos x+2)=0$

$\therefore \cos x-1=0$

$\cos x=1$

$x=0$ $\in$ $(-\pi ,\pi )$

$\therefore$ There are three roots in the given interval.

Hence, the answer is three.