Question

How do you solve `2sin^2x = 2 + cosx` in the interval 0 to 2pi?

Answer 1

First, perform a Pythagorean substitution to remove the sine term from the left side: `2(1-cos^2(x))=2 + cos(x)` .

Explanation:

Simplify the left side : `2-2cos^2(x)=2+cos(x)`
Gather like terms and set equal to 0: `0=2cos^2(x)+cos(x)`
Factor the right side: `0=cos(x)(2cos(x) + 1)`
Use the Zero Product Property:
`cos(x) = 0` or `2cos(x)+1=0`
`cos(x)=0` or `2cos(x) = -1`
`cos(x)=0` or `cos(x) = -1/2`
So, `x = cos^-1(0)` or `x = cos^-1(-1/2)`
x = `pi/2` , `(3*pi)/2`, or `(2*pi)/3`, `(4*pi)/3`