# Trig Equation

## Homework Statement

Solve for x: 2sinx-sin2x=4/∏

sin2x=2sinxcosx

## The Attempt at a Solution

2sinx-sin2x=4/∏

2sinx-2sinxcosx=4/∏

2sinx(1-cosx)=4/∏

sinx(1-cosx)=2/∏

And I get stuck here. Any suggestions or mistakes that I made? Thanks in advance.

## The Attempt at a Solution

Square [sinx(1-cosx)=2/∏], and modify the expression from sines to coses.
You'll end up having a quartic equation with two real solutions.

Square [sinx(1-cosx)=2/∏], and modify the expression from sines to coses.
You'll end up having a quartic equation with two real solutions.
So:

sinx(1-cosx)=2/∏

sin²x(1-cosx)²=4/∏ ²

sin² x(1-2cosx+cos² x)=4/∏ ²

(1-cos² x)(1-2cosx+cos² x)=4/∏ ²

1-2cosx+cos² x-cos² x+2cos³ x-cos^4x=4/∏ ²

1-2cosx+2cos^3x-cos^4x=4/∏ ²

Sorry but now I'm stuck here. Thanks for you help so far but can you give me another hint how to get past this part? Thanks.