I am struggling to know how to solve this equation as it involves more than one type of trigonometric function, I know how to do it with one repeated function. If a solution could be explained, that would be helpful as I can easily calculate the answer without understanding it.
Question:
$$4\sin^2{x} + 9\cos{x} - 6 = 0$$
for $0 < x < 720˚$
Hint Use $\sin^2 x = 1 - \cos^2 x$ to get:
$$4 - 4\cos^2 x + 9\cos x - 6 = 0$$
Use the substitution $u = \cos x$ and solve the quadratic equation.