We are given the equation:

$-2 \cos^2(x) = \cos(x) - 1$

Step 1: Move all terms to one side of the equation

First, let's rearrange the equation by moving all terms to one side:

$-2 \cos^2(x) - \cos(x) + 1 = 0$

Step 2: Solve the quadratic equation

We now have a quadratic equation in terms of $\cos(x)$. Let's rewrite it as:

$2 \cos^2(x) + \cos(x) - 1 = 0$

This is a quadratic equation in terms of $y = \cos(x)$, so we can solve it as we would any quadratic equation.

Step 3: Solve for $\cos(x)$ using the quadratic formula

The quadratic formula is given by:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $2y^2 + y - 1 = 0$, the coefficients are:

• $a = 2$
• $b = 1$
• $c = -1$

Substitute these values into the quadratic formula:

$y = \frac{-1 \pm \sqrt{(1)^2 - 4(2)(-1)}}{2(2)}$ $y = \frac{-1 \pm \sqrt{1 + 8}}{4}$ $y = \frac{-1 \pm \sqrt{9}}{4}$ $y = \frac{-1 \pm 3}{4}$

Thus, the two solutions for $y$ are:

$y_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$ $y_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$

Step 4: Solve for $x$

Now that we know $\cos(x) = \frac{1}{2}$ or $\cos(x) = -1$, we can solve for $x$ in the interval $[0, 2\pi)$.

Case 1: $\cos(x) = \frac{1}{2}$

The solutions for $\cos(x) = \frac{1}{2}$ in the interval $[0, 2\pi)$ are:

$x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3}$

Case 2: $\cos(x) = -1$

The solution for $\cos(x) = -1$ in the interval $[0, 2\pi)$ is:

$x = \pi$

Step 5: Final solutions

Thus, the exact solutions over the interval $[0, 2\pi)$ are:

$x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3}, \quad x = \pi$

Summary of solutions:

$x = \pi, \, \frac{\pi}{3}, \, \frac{5\pi}{3}$

Would you like further details on any step? Here are some additional questions for you to consider:

1. How can you solve for $x$ in different trigonometric equations, such as $\sin(x) = 1$?
2. How do you use the unit circle to find solutions to trigonometric equations?
3. How can the quadratic formula be used to solve more complex trigonometric equations?
4. Can this equation be solved using identities instead of the quadratic formula? If so, how?
5. How do you find solutions over different intervals, such as $[0, 360^\circ]$ or $[0, 2\pi]$?

Tip: For solving trigonometric equations, always try to rewrite the equation in terms of a single trigonometric function, such as $\sin(x)$, $\cos(x)$, or $\tan(x)$, to simplify the problem.