Solving Trig Equations

**watp** · December 28th 2010, 05:22 AM

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.

$2cos(2x)-3sin(x)-1=0$

Can anyone point me in the right direction?

**Prove It** · December 28th 2010, 05:26 AM

Use the identity $\displaystyle \cos{2x} = 1 - 2\sin^2{x}$ .

Then rearrange to get a quadratic equation in $\displaystyle \sin{x}$ that you can solve.

**watp** · December 28th 2010, 05:52 AM

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 $b^2-4ac$ ?

**e^(i*pi)** · December 28th 2010, 05:58 AM

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

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