Question

Please solve q18 ?

Answer 1

`rarrcos^6x+8cos^2x=4+4cos^4x`

Explanation:

`rarrsinx+sin^2x+sin^3x=1`

`rarrsinx(1+sin^2x)=1-sin^2x=cos^2x`

`rarrsin^2x(1+2sin^2x+sin^4x)=cos^4x`

`rarrsin^2x(1+2sin^2x+(1-cos^2x)^2)=cos^4x`

`rarrsin^2x(1+2(1-cos^2x)+1-2cos^2x+cos^4x)=cos^4x`

`rarrsin^2x(1+2-2cos^2x+1-2cos^2x+cos^4x)=cos^4x`

`rarr(1-cos^2x)(4-4cos^2x+cos^4x)=cos^4x`

`rarr4-4cos^2xcancel(+cos^4x)-4cos^2x+4cos^4x-cos^6x=cancelcos^4x`

`rarrcos^6x+8cos^2x=4+4cos^4x`