Question

If ${\sin }^{2}x+\sin x-1=0$, then the value of ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x$ is

• A. $1$
• B. $8$
• C. $-1$
• D. $-8$

Answer 1

The correct option is A $1$

Let $I={\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x$
$\Rightarrow I=({\cos }^{4}x{)}^3+3{\cos }^{6}x({\cos }^{4}x+{\cos }^{2}x)+({\cos }^{2}x{)}^3\Rightarrow I=({\cos }^{4}x+{\cos }^{2}x{)}^3$

Now, ${\sin }^{2}x+\sin x-1=0$
$\Rightarrow \sin x=1-{\sin }^{2}x\Rightarrow \sin x={\cos }^{2}x\cdots \left(1\right)\Rightarrow {\sin }^{2}x={\cos }^{4}x\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$I=({\sin }^{2}x+\sin x{)}^3\therefore I=1$