Posted by Alex on Sunday, March 23, 2008 at 4:52pm.
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
Not dividing both sides by cos x, subtracting cos x from the left side first, then dividing by cos x. Would that work?
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
sin x = -1/2 but I see those as x = pi+pi/6 = 7 pi/6 and at 2pi-pi/6 = 11pi/6
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
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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