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 pipi/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 2pipi/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
(x3) (x1) = 0
satisfied when x3 = 0 , so x = 3
and when x1 = 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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