Uber trig question, crazy problem with three roots.?

Expand sin(2x):

2sin(x)cos(x) + sin(x) + 2cos(x) + 1 = 0

2cos(x)(sin(x) + 1) + sin(x) + 1 = 0

(2cos(x) + 1)(sin(x) + 1) = 0

2cos(x) + 1 = 0

cos(x) = -1/2

x = arccos(-1/2)

x = 2pi/3, 4pi/3

sin(x) + 1 = 0

sin(x) = -1

x = arcsin(-1)

x = 3pi/2 (the third root)

ADD

ahahahahahahahaa ^^^ he changed his answer twice - initially gave four solutions, then said there are only two solutions but edited second time after seeing my answer... LOL

sin(2x) + sin(x) + 2cos(x) + 1 = 0.

2sin(x)cos(x) + sin(x) + 2cos(x) + 1 = 0.

sin(x)*(2cos(x) + 1) = -1 - 2cos(x). Square both sides.

sin^2(x)*(4cos^2(x) + 4os(x) + 1) = 1 + 4cos(x) + 4cos^2(x).

Let u = sin(x) and y = 1 + 4cos(x) + 4cos^2(x).

So we have u^2*y = y.

u^2*y - y = 0.

y(u^2 - 1) = 0.

y = 0 and u^2 - 1 = 0.

4cos^2(x) + 4cos(x) + 1 = 0. Take the square root of both sides.

2cos(x) + 1 = 0. ----> cos(x) = -1/2. x = 2pi/3 and x = 4pi/3.

And sin^2(x) - 1= 0. sin(x) = 1 and sin(x) = -1. x = pi/2 and x = 3pi/2. (But by graphing the equation, you can tell that pi/2 is an extraneous root; they don't solve the equation. So don't include it in the solution set.)

Solutions: x = 2pi/3, x = 4pi/3, and x = 3pi/2.

Hope this helps.

2.2