Question

`cosx+sinx=sqrt(cosx)`?

Answer 1

`rarrx=2npi` where `n in ZZ`

Explanation:

`rarrcosx+sinx=sqrtcosx`

`rarrcosx-sqrtcosx=-sinx`

`rarr(cosx-sqrtcosx)^2=(-sinx)^2`

`rarrcos^2x-2cosx*sqrtcosx+cosx=sin^2x=1-cos^2x`

`rarr2cos^2x-2cosx*sqrtcosx+cosx-1=0`

Let `sqrtcosx=y` then `cosx=y^2`

`rarr2*(y^2)^2-2*y^2*y+y^2-1=0`

`rarr2y^4-2y^3+y^2-1=0`

`rarr2y^3(y-1)+(y+1)*(y-1)=0`

`rarr[y-1][2y^3+y+1]=0`

Taking , `rarry-1=0`

`rarrsqrtcosx=1`

`rarrcosx=1=cos0`

`rarrx=2npi+-0=2npi` where `n in ZZ` which is the general

solution for `x`.