Posted by **Alex** on Sunday, March 23, 2008 at 4:52pm.

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?

- trig -
**Damon**, Sunday, March 23, 2008 at 5:22pm
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

- trig -
**Alex**, Sunday, March 23, 2008 at 5:35pm
Not dividing both sides by cos x, subtracting cos x from the left side first, then dividing by cos x. Would that work?

- trig -
**Damon**, Sunday, March 23, 2008 at 5:49pm
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

- typo 2 pi - pi/6 -
**Damon**, Sunday, March 23, 2008 at 5:52pm
sin x = -1/2 but I see those as x = pi+pi/6 = 7 pi/6 and at 2pi-pi/6 = 11pi/6

- trig -
**Damon**, Sunday, March 23, 2008 at 5:56pm
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

- trig -
**Alex**, Sunday, March 23, 2008 at 6:11pm
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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