Question

How do you solve for x in `3sin2x=cos2x` for the interval `0 ≤ x < 2π`

Answer 1

You can use the fact that `tan(x)=sin(x)/cos(x)`
In your expression.:
`3*sin(2x)=cos(2x)`
`3*sin(2x)/cos(2x)=1` and
`tan(2x)=1/3`
`tan(2x)` is equal to `1/3` for `2x=0.321 and 3.463` rad in your interval;

(Graph from: http://www.intmath.com/trigonometric-graphs/4-graphs-tangent-cotangent-secant-cosecant.php)

and so to have `x` you divide by `2` to get: `x=0.160 and 1.731` rad

Hope it helps