# Trigonometric equation

## Homework Statement

2 sin(2x) = 2 cos(x)

## The Attempt at a Solution

4sinxcosx-2cosx=0

2(2sinxcosx-cosx)=0

2cosx(sinx-1)=0

2cosx=0 sinx=1

And then I did the math and got pi/2 and 3pi/2 , but that's apparently wrong.

## The Attempt at a Solution

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jedishrfu
Mentor
I get cosx * (2*sinx - 1) = 0 for your second to last step. Does that help?

I still get pi/2 and 3 pi/2.

>_<

I'll try consulting my notes on how to properly calculate cosx=0 and sinx=1/2.

vela
Staff Emeritus
Homework Helper
I still get pi/2 and 3 pi/2.
How? You know that $\sin (\pi/2) = 1$ and $\sin (3\pi/2) = -1$. So obviously neither of those satisfy $\sin x = 1/2$.

arcsin (1/2) = x1
arccos (0) = x2

You have done everything correctly, now you need to figure out what the values for x are. This is not a linear equation, there are 2 series of solutions.