Find ALL solutions of the interval [0, 2π)

Question

Find ALL solutions of the interval [0, 2π)

cos x/2 = - √3/2

If there are solutions that will not fit, please explain.

Thank you!

Answer 1

cosx/2 =0 sqr3/2

x/2 = arccos-sqr3/2=5pi/6

x =10pi/6= 5pi/3

Answer 2

Find all of the solutions of cox (x/2) = -√3/2 on [0,2π)

First we need to find where cos(x) = -√3/2. To do this, we can find the reference angle by determining the angle where cox(x)=√3/2 either using the first quadrant of the unit circle or by looking at the ratios of a 30-60-90 right triangle.

Our reference angle is π/6

Since we need our cosine value to be negative, we need to place that reference angle in the second and third quadrants of the unit circle. To find the angle in the second quadrant, we subtract the reference angle from π, and to find the angle in the third quadrant we add the reference angle to π.

π - π/6 = 5π/6

π + π/6 = 7π/6

This means that

x/2 = 5π/6 , 7π/6

But, because trig functions are cyclical, we would hit these same locations every time we spin around the unit circle. To show that, we would write

x/2 = (5π/6 + 2πk) , (7π/6 + 2πk) where k = 0, 1 , 2, 3 ,4 . . .

To solve for x, we would multiply both sides by 2, distributing to all values on the right.

x = (5π/3 + πk), (7π/3 + πk)

If we only want the values on the domain [0,2π) we can write a list of answers using our k values (starting with k=0).

x = 5π/3 , 7π/3, 8π/3, 10π/3 . . .

We can see that our x values quickly become too large to fit in the given domain. So our final answer is

x = 5π/3

Answer 3

cos(x/2) = -√3/2

Cos^-1(√3/2) = π/6 = reference angle

Cosine is negative in quadrants 2 and 3.

So, x/2 = 5π/6 + 2kπ, or 7π/6 + 2kπ, where k = 0, 1, -1, 2, -2,...

x = 5π/3 + 4kπ, or 7π/3 + 4kπ

The only solution in the interval [0, 2π) is 5π/3.