# Solve 2cos^{2}x + cosx - 1 = 0 for x over the interval [0, 2π).

**Solution:**

Given, equation is 2cos^{2}x + cosx - 1 = 0

We have to solve for x over the interval [0, 2].

Now, 2cos^{2}x + cosx - 1 can be written as 2cos^{2}x + 2cosx - cosx - 1

So, 2cos^{2}x + 2cosx - cosx - 1 = 0

By grouping,

2cos^{2}x + 2cosx - 1(cosx + 1) = 0

2cosx(cosx + 1) - 1(cosx + 1) = 0

(2cosx - 1)(cosx + 1) = 0

Now, 2cosx - 1 = 0

2cosx = 1

cosx = 1/2

x = cos^{-1}(1/2)

x = π/3, 5π/3

Now, cosx + 1 = 0

cosx = -1

x = cos^{-1}(-1)

x = π

Therefore, the value of x over the given interval are x = π, π/3 and 5π/3.

## Solve 2cos^{2}x + cosx - 1 = 0 for x over the interval [0, 2π).

**Summary:**

Solving 2cos^{2}x + cosx - 1 = 0 for x over the interval [0, 2], the value of x are π, π/3 and 5π/3.

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