Im stuck on this question in my maths homework:

Solve the equation √3tan2x-1=0 for 0<X<π

Here is what i've done:

√3tan2x-1=0

tan2x=1/√3

Using trig identities,

tanx= sinx/cosx

so tan2x=sin2x/cos2x < is this right?

I've then used the trig expansions:

2sinxcosx/1-2sin^2x < Am I using the right expansion for cos2x?

Am I using the right method? I have tried to go further than this, but nothing has worked out, so where do I go from here?

Thanks in advance.