Question

How do you solve `2cos^2x+sinx-2=0` over the interval 0 to 2pi?

Answer 1

I found:
`x=0, pi,2pi`
`x=pi/6,5/6pi`

Explanation:

We can use the fact that:
`cos^2(x)=1-sin^2(x)`
and write:
`2(1-sin^2(x))+sin(x)-2=0`
`cancel(2)-2sin^2(x)+sin(x)cancel(-2)=0`
solve as a normal Quadratic Equation but in `sin(x)` instead of `x`:
collecting `sin(x)`:
`sin(x)[-2sin(x)+1]=0`
we get 2 solutions:
`sin(x)=0`
and:
`-2sin(x)+1=0`
so that must be:
`sin(x)=0` when `x=0, pi,2pi`
`sin(x)=1/2` when `x=pi/6,5/6pi`