# Pre-Cal(Urgent!!)

posted by .

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!!) -

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!!) -

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!!) -

I agree with Reiny up until the 11 pi/6

• Pre-Cal(Urgent!!) -

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)

## Similar Questions

1. ### math

find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 x Since sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin …
2. ### Pre-Calculus

cos^2(x) + sin(x) = 1 - Find all solutions in the interval [0, 2pi) I got pi/2 but the answer says {0, pi/2, pi} Where does 0 and pi come from for solutions?
3. ### Pre-Cal

Find all the soultions of the equation in the interval (0,2pi) sin 2x = -sqrt3 /2 sin -sqrt3 /2 is 4pi/3 and 5pi/3 in the unit circle 2x= 4pi/3 + 2npi 2x=5pi/3 + 2npi I do not know what to do at this point.

Find all the soultions of the equation in the interval (0,2pi) sin 2x = -sqrt3 /2 sin -sqrt3 /2 is 4pi/3 and 5pi/3 in the unit circle 2x= 4pi/3 + 2npi 2x=5pi/3 + 2npi I do not know what to do at this point

1) Find all the solutions of the equation in the interval (0,2pi). sin 2x = -sqrt3/2 I know that the angles are 4pi/3 and 5pi/3 and then since it is sin I add 2npi to them. 2x = 4pi/6 + 2npi 2x = 5pi/3 + 2npi Now I know that I have …

1) Find all the solutions of the equation in the interval (0,2pi). sin 2x = -sqrt3/2 I know that the angles are 4pi/3 and 5pi/3 and then since it is sin I add 2npi to them. 2x = 4pi/6 + 2npi 2x = 5pi/3 + 2npi Now I know that I have …
7. ### Pre-Cal

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.
8. ### Pre-Cal

Approximate the equation's soultions in the interval (o, 2pi). If possible find the exact solutions. sin 2x sinx = cosx I do not know where to start.