If sin x + sin

We have:

$\sin x+\sin ^{2} x=1$       .....(1)

$\Rightarrow \sin x=1-\sin ^{2} x$

$\Rightarrow \sin x=\cos ^{2} x$      ..(2)

Now, taking cube of (1):

$\sin x+\sin ^{2} x=1$

$\Rightarrow\left(\sin x+\sin ^{2} x\right)^{3}=(1)^{3}$

$\Rightarrow(\sin x)^{3}+\left(\sin ^{2} x\right)^{3}+3(\sin x)^{2}\left(\sin ^{2} x\right)+3(\sin x)\left(\sin ^{2} x\right)^{2}=1$

$\Rightarrow(\sin x)^{3}+(\sin x)^{6}+3(\sin x)^{4}+3(\sin x)^{5}=1$

$\Rightarrow(\sin x)^{6}+3(\sin x)^{5}+3(\sin x)^{4}+(\sin x)^{3}=1$

$\Rightarrow\left(\cos ^{2} x\right)^{6}+3\left(\cos ^{2} x\right)^{5}+3\left(\cos ^{2} x\right)^{4}+\left(\cos ^{2} x\right)^{3}=1$

$\Rightarrow \cos ^{12} x+3 \cos ^{10} x+3 \cos ^{8} x+\cos ^{6} x=1$