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.
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)
To solve the equation sin(x + pi/6) - sin(x -pi/6) = 1/2 in the interval (0, 2pi), let's break it down step by step:
1. Use the sine difference identity to simplify:
sin(a - b) = sin(a)cos(b) - cos(a)sin(b)
Applying this identity to our equation:
sin(x)cos(pi/6) - cos(x)sin(pi/6) - sin(x)cos(-pi/6) + cos(x)sin(-pi/6) = 1/2
Simplifying further:
(1/2)sin(x) - (sqrt(3)/2)cos(x) - (-1/2)sin(x) - (sqrt(3)/2)cos(x) = 1/2
Combining like terms:
sin(x) - sqrt(3)cos(x) - sin(x) - sqrt(3)cos(x) = 1/2
Simplifying again:
-2sqrt(3)cos(x) = 1/2
2. Divide both sides of the equation by -2sqrt(3):
cos(x) = -1/(4sqrt(3))
3. Use the cosine function table or calculator to find the angles whose cosine is equal to -1/(4sqrt(3)) in the interval (0, 2pi).
On the unit circle, cosine is negative in the second and third quadrants. Looking at the principal values for cosine 1/(4sqrt(3)),
we find that cos(pi/6) = sqrt(3)/2 and cos(5pi/6) = -sqrt(3)/2.
Therefore, the angles that satisfy cos(x) = -1/(4sqrt(3)) are pi/6 and 5pi/6.
4. Determine the values of x in the interval (0, 2pi) that correspond to the solutions found in Step 3.
Considering these values:
x = pi/6 and x = 5pi/6
However, we need to ensure that these solutions fall within the given interval (0, 2pi).
Checking for pi/6:
0 < pi/6 < 2pi - This solution is valid.
Checking for 5pi/6:
0 < 5pi/6 < 2pi - This solution is valid.
Therefore, the solutions in the interval (0, 2pi) are x = pi/6 and x = 5pi/6.