Solve The Equation in the Interval [0,2pi)?

Question

2cosx-3tanx=0

My previous exercise asked Sinx-2Sin^2x=0, which I could do fine-- But I don't understand this one. If someone can help, I'd appreciate it!

Thanks Ron, that helps a lot. But one question, isn't it, when it's at "2Cos^2x-3Sinx" actually "(2Cos^2x-3Sinx)/Cosx"?

Answer 1

tan(x) = sin(x)/cos(x)

Multiply both sides by cos(x):

2cos²(x) - 3 sin(x) = 0

We'd like to get this all in either sines or cosines. Since cos²(x) = 1 - sin²(x), that will do the trick:

2(1 - sin²(x)) - 3 sin(x) = 0

This is a quadratic equation for sin(x). You will have to discard one of the roots, -2, because -1 ≤ sin(x) ≤ 1.