
Solving Trig Equations
I have a trig equation that has both sin and cos in it, and I have no clue how to solve it.
I know how to solve it with one (move the numbers to the right, do the inverse), but not with two.
$\displaystyle 2cos(2x)3sin(x)1=0$
Can anyone point me in the right direction?

Use the identity $\displaystyle \displaystyle \cos{2x} = 1  2\sin^2{x}$.
Then rearrange to get a quadratic equation in $\displaystyle \displaystyle \sin{x}$ that you can solve.

Thanks for the speedy reply.
I'm not quite sure what you mean. Where does the square come from in sin x? By quadratic equation, do you mean $\displaystyle b^24ac$?

That's the discriminant. A quadratic equation is of the form $\displaystyle ax^2+bx+c=0$