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.
To solve the equation sin(x + pi/6) - sin(x -pi/6) = 1/2, you correctly simplified the expression on the left-hand side to 2sin(x)(sqrt(3)/2) = 1/2.
First, let's solve this equation: 2sin(x)(sqrt(3)/2) = 1/2.
To solve for sin(x), divide both sides by 2(sqrt(3)/2):
sin(x) = (1/2) / [(sqrt(3)/2)]
Next, simplify the expression on the right-hand side by multiplying the numerator and denominator by the conjugate of the denominator:
sin(x) = (1/2) * [(2/sqrt(3))/[(2/sqrt(3))]]
Now, we can simplify further:
sin(x) = (1/2) * (2/sqrt(3)) * (sqrt(3)/sqrt(3))
This simplifies to:
sin(x) = (1/2) * (2sqrt(3))/(3)
Finally, simplify further:
sin(x) = (sqrt(3))/(3)
Now, to find the solutions of sin(x) = (sqrt(3))/(3) in the given interval (0, 2pi), we need to find the angles whose sine is equal to (sqrt(3))/(3).
The principal value of the angle whose sine is (sqrt(3))/(3) is pi/3. Since sine is a periodic function with a period of 2pi, we can find other solutions in the interval (0, 2pi) by adding or subtracting integer multiples of the period (2pi) to the principal value.
So, the solutions in the given interval are:
x = pi/3, 4pi/3