# 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?

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

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

posted by Alex
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

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

posted by Damon
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

posted by Damon
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.

posted by Alex

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