Question

Solve 6 cos^2(x) + cos(x) - 1 = 0 for all solutions 0 < x < 2π.
x=

Answer 1

The solution to the trigonometry equation in the interval 0 < x < 2π is x = 2π/3 and 4π/3

How to determine the solution to the trigonometry equation

From the question, we have the following parameters that can be used in our computation:

6cos²(x) + cos(x) - 1 = 0

Let y = cos(x)

So, we have

6y² + y - 1 = 0

Expand the equation

6y² + 3y - 2y - 1 = 0

Factorize

3y(2y + 1) - 1(2y + 1) = 0

So, we have

(3y - 1)(2y + 1) = 0

Expand

3y - 1 = 0 and 2y + 1 = 0

Evaluate

y = 1/3 and y = -1/2

Recall that

y = cos(x)

So, we have

cos(x) = 1/3 and cos(x) = -1/2

Take the arccos and evaluate

x = 2π/3 and 4π/3

Hence, the solution to the trigonometry equation is x = 2π/3 and 4π/3