Solving Trigonometric equations/Summer HW?

Question

Hey!
I was wondering if anyne can help me solve this equation:
2sinxcosx+sinx=0 ??

I seem to have become a little rusty over the summer

I should have mentioned this above, but it has to be on the interval [0,2pi)

Thanks to all who helped out!

Answer 1

2sinx * cosx + sin x = 0
sinx (2cosx + 1) = 0
sinx = 0 or 2cosx + 1 = 0
x = k * Pi or cosx = -1/2
x = k * PI or x = 2 * PI/3 + 2*k'*Pi or x = -2*Pi/3 + 2k"*Pi

Answer 2

Let's find all x with 0 <= x < 2π that satisfy this equation.
Factoring the left side, we get
sin x(2 cos x + 1) = 0.
Now set each factor equal to zero and solve:
sin x = 0, so x = 0 and π
2 cos x + 1 = 0
cos x = -1/2.
The reference angle is π/3 and cos x is negative.
That means x is in the second and third quadrants.
So x = 2π/3 and x = 4π/3.
Hope that helps!

Answer 3

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

Therefore:
sin(x) = 0
x = 0, pi

or

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

Answer 4

2sinxcosx+sinx=0
2 sin x cos x + sin x = 0
---------------------------------
cos x

2 sin x + tan x = 0

2 sin x = tan x
2 sin x = (sin x)/(cos x)
/2

sin x = sin x/(2 cos x)