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