Question

3sin thita = 4 cos thita
4sin square- 3cos+2?

Answer 1

theta, not thita

Nice equations -- are you supposed to solve them?

Recall that tanθ = sinθ/cosθ
3sinθ = 4cosθ
tanθ = 4/3
θ = 53.13°

For the second one, is that your attempt to solve the first line? If not,
Recall that sin^2θ + cos^2θ = 1
4sin^2θ - 3cosθ + 2 (I think - typos mess things up)
No equation here, but if you meant
4sin^2θ - 3cosθ + 2 = 0
4(1-cos^2θ)-3cosθ+2 = 0
4-4cos^2θ - 3cosθ + 2 = 0
4cos^2θ + 3cosθ - 6 = 0
Now it's just a quadratic to solve.