# Trigonometric equation

GuN

## 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.

## Answers and Replies

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

GuN
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.

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##.

lendav_rott
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.

The system I showed you will not be your final answer.

GuN
Right, I double checked the notes and found I used the wrong radians.

I redid it and got (for the domain of [0, 2pi] ) pi/6, 5pi/6, pi/2 and 3pi/2.