Posted by **Abby** on Saturday, February 20, 2010 at 12:54pm.

Find all the solutions of the equation in the interval (0,2pip)

sin(x + pi/6) - sin(x -pi/6) = 1/2

I am only stuck on the last part. I have 2sinx(sqrt3/2) = 1/2 Does the 2 cancel out with the sqrt 3/2. I am not sure what to do.

- Pre-Cal(Urgent!!) -
**Reiny**, Saturday, February 20, 2010 at 1:09pm
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

- Pre-Cal(Urgent!!) -
**drwls**, Saturday, February 20, 2010 at 1:15pm
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

- Pre-Cal(Urgent!!) -
**drwls**, Saturday, February 20, 2010 at 1:16pm
I agree with Reiny up until the 11 pi/6

- Pre-Cal(Urgent!!) -
**Reiny**, Saturday, February 20, 2010 at 3:26pm
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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