Question

If $\sin x+{\sin }^{2}x+{\sin }^{3}x=1$, then ${\cos }^{6}x-4{\cos }^{4}x+8{\cos }^{2}x$ is equal to

• A. $0$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

$\sin x+{\sin }^{2}x+{\sin }^{3}x=1\Rightarrow \sin x+{\sin }^{3}x=1-{\sin }^{2}x\Rightarrow \sin x(1+{\sin }^{2}x)={\cos }^{2}x\Rightarrow \sin x(2-{\cos }^{2}x)={\cos }^{2}x\Rightarrow {\sin }^{2}x{(2-{\cos }^{2}x)}^2={\cos }^{4}x\Rightarrow (1-{\cos }^{2}x)(4+{\cos }^{4}x-4{\cos }^{2}x)={\cos }^{4}x\Rightarrow 4+{\cos }^{4}x-4{\cos }^{2}x-4{\cos }^{2}x-{\cos }^{6}x+4{\cos }^{4}x={\cos }^{4}x\Rightarrow {\cos }^{6}x-4{\cos }^{4}x+8{\cos }^{2}x=4$