How do I solve equations like 3sin^2(x) - 2cos(x) = 2

To solve equations like this we need to get the equation in terms of one trigonometric function, ie one of sin(x), cos(x) or tan(x)

It is usually easiest to get everything in terms of whatever the linear part of the equation is, in this case cos(x).

To do this we use that sin²(x) + cos²(x) = 1. Since we're trying to get rid of the sin²(x) part of the equation we rearrange this to get that sin²(x) = 1 - cos²(x)

We substitute this into the equation 3sin²(x) - 2cos(x) = 2

This gives 3(1 - cos²(x)) - 2cos(x) = 2

This then becomes 3 - 3cos²(x) - 2cos(x) = 2

which we can rearrange to 3cos²(x) + 2cos(x) - 1 = 0

Now that we have an equation just in terms of one trigonometric function , here cos(x), we can just replace it with some letter, say y

Thus this is just 3y² + 2y - 1 = 0. We can solve this with the usual quadratic equation, giving y = -1 and y = 1/3. As y = cos(x) this is just cos(x) = -1 and cos(x) = 1/3. To find x we then just solve these in the usual way