Question

Solve `2cos^2x+sinx=1` ?

Answer 1

`x = (pi/2+2k pi uu -pi/6 +2kpi)` for `k in NN`

Explanation:

Using the identity

`sin^2x+cos^2x=1` and substituting

`2(1-sin^2x)+sinx-1=0` or

`2sin^2x-sinx-1=0`

now solving for `sinx`

`sinx = (1pm sqrt(1+8))/4 = {(1),(-1/2):}`

then

`x = (pi/2+2k pi uu -pi/6 +2kpi)` for `k in NN`