Can someone tell me how to begin this problem? I know I have to use a double angle formula: sin(2x)-sinx=(2cos^2)x-cosx, find all solutions in the interval [0,2pi)

Question

zepdrix · Answer

Yah this one is a little tricky, this is what I would do... After applying the Sine Double Angle to the first term, we have,$$\large 2\sin x\cdot \cos x-\sin x\quad =\quad 2\cos^2x-\cos x$$From there we can factor out a sine from each term on the left, a cosine from each term on the right, giving us,

zepdrix · Answer

$$\large \sin x(2\cos x-1)\quad = \quad \cos x(2\cos x - 1)$$See where this is going? :)

zehanz · Answer

Begin with setting sin2x = 2sinxcosx and then look for common factors left and commmon factors right. See zepdrix's work!

zepdrix · Answer

Oh looks like he logged off :D lol

zehanz · Answer

At least we had some fun ;)

anonymous · Answer

Thank you guys!