#1**0 **

sin2x-sinx=0

Let's use an identity for sin(2x)......this = 2 sinxcosx

So we have

2sinxcosx - sinx = 0........let's factor out the sin x....so we have

sinx (2cosx -1) = 0

Setting each factor to 0, we have

sinx = 0 and 2cosx -1 =0

Solving the first, we have x = 0 + (pi)n

Solving the second, we have

cosx = 1/2

so x = (-pi/6) + 2(pi)n

and x = (pi/6) + 2(pi)n

Chris Apr 14, 2014

#2**0 **

Oops!!!!!...I made a slight mistake in this problem in the second part.......(sorry)...let me correct thiis...

I wrote

cosx = 1/2

so x = (-pi/6) + 2(pi)n

and x = (pi/6) + 2(pi)n

This* should* be

cosx = 1/2

so x = (-pi/3) + 2(pi)n

and x = (pi/3) + 2(pi)n

(I should know better than to try to do math late at night!!!)

Chris Apr 14, 2014