If `sinx+sin^2x+sin^3x=1` then find the value of `cos^6x-4cos^4x+8cos^2x`

Explanation

We have ,
`sinx+sin^(2)x+sin^(3)x=1`
`rArrsinx+sin^(3)x=1-sin^(2)x`
`rArrsinx(1+sin^(2)x)=cos^(2)x`
`rArrsinx(1+1-cos^(2)x)=cos^(2)x`
`rArrsinx(2-cos^(2)x)=cos^(2)x`
Squaring both sides ,we get
`sin^(2)x(2-cos^(2)x)^(2)=cos^(4)x`
`rArr(1-cos^(2)x)(4+cos^(4)x-4cos^(2)x)=cos^(4)x`
`rArr4+cos^(4)x-4cos^(2)x-4cos^(2)x-cos^(6)x+4cos^(4)x=cos^(4)x`
`rArrcos^(6)x-4cos^(4)x+8cos^(2)x-4=0`