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

  • Thread starter Thread starter maff is tuff
  • Start date Start date
AI Thread 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.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
View full post »

Similar threads

How Many Solutions Does the Trigonometric Equation Have in the Given Interval?
Replies
5
Views
2K
Double Angle Trig: Solving Sin2x-cosx=1 for x in [0,2pi)
Replies
5
Views
2K
Transforming sinx to sinx + cosx
Replies
6
Views
2K
Loosing solutions in a trig equation
Replies
6
Views
2K
How to Solve Trig Identity Questions: Tips and Examples
Replies
28
Views
4K
Proving Reciprocal Identities: (secx+1)/(sin2x) = (tanx)/2cosx-2cos2x
Replies
10
Views
2K
Need help proving some trig identities
Replies
4
Views
2K
How Do You Solve Trigonometric Equations Involving Sin, Cos, and Tan?
Replies
7
Views
3K
Mastering Trig Equations: Solving for x in cosx(2sinx+1) = 0
Replies
3
Views
8K
Differential Equations: Solve the following
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top