How would you go about finding all the solutions of the equation. I just don't know where to start.
2cos2theta-root3=0
don't forget your algebra I just because you're doing trig now.
2cos2θ = √3
cos2θ = √3/2
Now it's time to recall where cos(x) is positive: QI and QIV
2θ = π/6 or -π/6
Now, cos(x) has period 2π, so cos(2x) has period π, since 2x changes twice as fast as x.
So, now we have θ = π/12 or -π/12.
But any multiple of π will also work. So,
θ = kπ ± π/12
see the graph at
http://www.wolframalpha.com/input/?i=plot+2cos2%CE%B8+%3D+%E2%88%9A3+for+-2pi+%3C%3D+%CE%B8+%3C%3D+2pi