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

I'm looking for exact value solutions in [0, 3π]

So I need to find general solutions to solve the equation. But do I eliminate cos(x), like this...

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

2sin(x)cos(x)= -cos(x)

2sin(x) = -1

sin(x) = -1/2 at 4pi/3 and 5pi/3

and then use those general solutions to find my exact values? Or do I need to incorporate the cos(x) somehow, by factoring it out of the original equation like...

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

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

and then use general solutions for that?

when you divide both sides by cos x, you divide by zero when cos x = 0. Not allowed.

However your equation is satisfied if cos x = 0, so at x = pi/2 and at x = 3 pi/2
Your other reasoning is also valid
sin x = -1/2 but I see those as x = pi+pi/6 = 7 pi/6 and at pi-pi/6 = 5 pi/6

Not dividing both sides by cos x, subtracting cos x from the left side first, then dividing by cos x. Would that work?

What you did later worked fine but dividing by cos x, before or after moving it to the right, is suspect in my mind.

However what you did later:
2sin(x)cos(x)+cos(x)=0
then
cos(x)[2sin(x)+1]=0
so zero when cos x = 0
or when sin x = -1/2
is just fine and safe to my mind.
I disagee with you about where in quadrants 3 and 4 sin x = -1/2

sin x = -1/2 but I see those as x = pi+pi/6 = 7 pi/6 and at 2pi-pi/6 = 11pi/6

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

so zero when cos x = 0
or when sin x = -1/2
note
this is just like factoring a quadratic
x^2 - 4x + 3 = 0
(x-3) (x-1) = 0
satisfied when x-3 = 0 , so x = 3
and when x-1 = 0 so x = 1

My mistake about the locations of where sin(x) is equal to -1/2... &#*@ now I have to redo two lengthy problems.

Thanks, though.

To find exact value solutions in the interval [0, 3π] for the equation 2sin(x)cos(x) + cos(x) = 0, you can approach it in different ways.

Method 1: Eliminate cos(x)
You correctly eliminated cos(x) from the equation and obtained the equation 2sin(x) = -1. To find the solutions within the given interval, you can use the fact that sin(x) = -1/2 at the angles 4π/3 and 5π/3. These are the exact value solutions for the equation in the given interval.

Method 2: Incorporate cos(x)
You can also work with the original equation 2sin(x)cos(x) + cos(x) = 0 by factoring out cos(x) like this: cos(x)[2sin(x) + 1] = 0. This equation holds true if either cos(x) = 0 or (2sin(x) + 1) = 0.

For cos(x) = 0, you can find the solutions within the interval [0, 3π] by considering the angles where cos(x) equals zero, which are x = π/2 and x = 3π/2.

For (2sin(x) + 1) = 0, you need to solve for sin(x) by rearranging the equation: 2sin(x) = -1, sin(x) = -1/2. Again, you can find the solutions for sin(x) = -1/2 within the interval [0, 3π] using the angles you mentioned: x = 4π/3 and x = 5π/3.

In summary, you can find the exact value solutions within the interval [0, 3π] by considering both cos(x) = 0 (at x = π/2 and x = 3π/2) and sin(x) = -1/2 (at x = 4π/3 and x = 5π/3).