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?

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 pipi/6 = 5 pi/6
posted by Damon

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

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 = 1/2posted by Damon

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

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
(x3) (x1) = 0
satisfied when x3 = 0 , so x = 3
and when x1 = 0 so x = 1posted by Damon

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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