Question

How do you solve #2sin^2x-sinx-1=0 and find all exact general solutions?

Answer 1

`x=(7pi)/6+2kpi or x=(11pi)/6+2kpi or x=pi/2+2kpi`

Explanation:

By using the quadratic formula with respect to sin x, you get:

`sinx=(1+-sqrt(1-4*2*(-1)))/(2*2)`

`sinx=(1+sqrt(9))/4=(1+-3)/4`

Then `sinx=-1/2 or sinx=1`

that's

`x=(7pi)/6+2kpi or x=(11pi)/6+2kpi or x=pi/2+2kpi`