Solving for x: Find the Solution to 2sinx-sin2x=4/∏

  • Thread starter Thread starter maff is tuff
  • Start date Start date
Click For Summary
The equation 2sinx - sin2x = 4/π can be transformed using the identity sin2x = 2sinxcosx, leading to the expression 2sinx(1 - cosx) = 4/π. This simplifies to sinx(1 - cosx) = 2/π. To progress, squaring both sides results in sin²x(1 - cosx)² = 4/π², which can be expanded and rearranged into a quartic equation. The discussion highlights that this quartic equation has two real solutions, but the user is seeking further assistance to navigate the complexities of the algebra involved. Additional hints are requested to help overcome the current impasse.
maff is tuff
Messages
65
Reaction score
1

Homework Statement



Solve for x: 2sinx-sin2x=4/∏

Homework Equations



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.
 
Physics news on Phys.org
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.
 
Atran said:
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.
 

Similar threads

Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 28 ·
Replies
28
Views
4K
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
3K
Replies
3
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K