I don't see how you got 2sinx(sqrt3/2) = 1/2 .
Here is what I would do:
sin(x + pi/6) - sin(x -pi/6) = 1/2
sinxcospi/6 + cosxsinpi/6 - (sinxcospi/6 - cosxsinpi/6) = 1/2
2cosxsinpi/6 = 1/2
cosx = 1/2, since sinpi/6 = 1/2
so x is in the first or fourth quadrants
In quad I, x = pi/3 , (60 degrees) or
in quad IV, x = 2pi - pi/6 = 11pi/6 , (300 degrees)
BTW, if 2sinx(sqrt3/2) = 1/2 were correct,
you could NOT cancel the 2's
You got a sign wrong somewhere.
Using the sin (a+b) and sin (a-b) formulas, the left half becomes
sinx cos pi/6 + sin(pi/6) cosx
- sinx cos(pi/6) + cosx sin(pi/6) = 1/2
2 cosx sin pi/6 = 1/2
cos x = 1/2
x = pi/3 or 5 pi/3
I agree with Reiny up until the 11 pi/6
Yup, drwls is right,
I should have had 2pi - pi/3,
x = 5pi/3,
(I had 2pi - pi/6, don't know where that came from)
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