Find the solutions to the equation in the interval 0 ≤ 𝛽 ≤ 𝜋 .

Question

2 cos(2𝛽) + 1 = 0

𝛽 = 

Answer 1

For all problems similar to this we want to first start off by isolating the variable and then solve from there.

2cos(2β) = -1

cos(2β) = -1/2

2β = arccos(-1/2)

β = (arccos(-1/2))/2

Now that we have isolated the variable, let us solve and find all the solutions to this equation.

Looking at the arccos(-1/2), when does cos(Θ) = -1/2

Using the unit circle this happens when either Θ = 2π/3 + 2πk, 4π/3 + 2πk

Since the arccos(-1/2) is divided by 2 we also have to divide what we get from the arccos by 2

This leaves us with Θ = π/3 + πk, 2π/3 + πk

Where k is some constant

Now that we have all the solutions, now find all the solutions that are within the interval

∴ β = π/3, 2π/3

Answer 2

cos(2β) = - 1/2; 2β = 2π/3 or 2β = 4π/3; β = π/3, or β = 2π/3.