# trig

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?

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1. 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

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

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3. 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

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

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5. 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

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

Thanks, though.

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