SOLUTION: solve for x for: sinx - sin2x = cosx on [-pi, pi]

Algebra.Com
Question 985071: solve for x for: sinx - sin2x = cosx on [-pi, pi]
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Is it ?
Or ?
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
sin(x) - sin(2x) = cos(x)

Any way you try this it always involves solving a 4th degree polynomial 
equation, so I'd just use a graphing calculator to solve it. 

Solutions:  -2.8084729373435, 1.2376766105486

Edwin
RELATED QUESTIONS

Solve: {{{sin2x=cosx}}},... (answered by Alan3354,solver91311)
15. Solve for 0≤𝑥<2𝜋: cosx ∙ sinx =... (answered by greenestamps)
show that sin x/ 1-cosx = 1+ cosx/sinx for every (answered by Boreal)
if u= int (f(sin2x)sinx)dx on [0 ,pi/2] and v= int (f(cos2x)cosx)dx on [0 ,pi/2] then... (answered by ikleyn)
How do you solve for x in cos2x = cosx,... (answered by robertb)
Solve the eq on the interval [0,2pi) : sin2x-cosx=0 I used the indetity sin2x =... (answered by solver91311,Alan3354)
Solve: sin2x = cos x ,{{{-2pi <= x <=... (answered by Alan3354)
cos^2x - sinx cosx solve for... (answered by Alan3354)
Hello. I am having difficulty with this question and I would ask for your help. Solve... (answered by lwsshak3)