Question

How do you find the critical numbers of `y=cos x + sin 2x`?

Answer 1

`y'=-sinx+2cos2x = -sinx+2(1-2sin^2x)`

`y' = -4sin^2x-sinx+2`

`y' = 0` at solutions to

`4sin^2x+sinx-2=0`

`sinx = (-1+-sqrt(1+32))/8 = (-1 +-sqrt33)/8`

`x= {(sin^-1(-1-sqrt33)/8 +2pik),(-pi-sin^-1(-1-sqrt33)/8 +2pik) ,(sin^-1(-1+sqrt33)/8 +2pik),(pi-sin^-1(-1+sqrt33)/8 +2pik):}` `" "` for integer `k`