Question
The sum of the real roots of $${\cos ^6}x + {\sin ^4}x = 1$$ in the interval $$ - \pi \leqslant x \leqslant \pi $$ is equal to
A.
$$0$$
B.
$$\pi $$
C.
$$ - \pi $$
D.
None of these
Answer :
$$0$$
Solution :
$$\eqalign{
& {\cos ^6}x = 1 - {\sin ^4}x = {\cos ^2}x\left( {1 + {{\sin }^2}x} \right) \cr
& \Rightarrow \,\,{\cos ^2}x = 0\,\,\,{\text{or, }}{\cos ^4}x = 1 + {\sin ^2}x \cr
& 1 + {\sin ^2}x = {\cos ^4}x \cr
& \Rightarrow \,\,{\sin ^2}x = - \left( {1 - {{\cos }^4}x} \right) = - {\sin ^2}x\left( {1 + {{\cos }^2}x} \right) \cr
& \Rightarrow \,\,{\sin ^2}x = 0\,\,\,{\text{or, }}2 + {\cos ^2}x = 0\left( {{\text{absurd}}} \right). \cr
& \therefore \,\,\cos x = 0\,\,\,{\text{or, }}\sin x = 0 \cr
& \Rightarrow \,\,x = \frac{\pi }{2}, - \frac{\pi }{2},0,\pi , - \pi . \cr} $$