Amanita, let's carefully solve the equation given in the green rectangle:

$2\cos^2{x} + 5\sin{x} = 4, \quad 0 \leq x \leq 2\pi$

Here’s the step-by-step plan:

Step 1: Express everything in terms of $\sin{x}$

We know that $\cos^2{x} = 1 - \sin^2{x}$. This substitution will help us turn the equation into a quadratic in terms of $\sin{x}$.

Step 2: Substitute and simplify

Replace $\cos^2{x}$ with $1 - \sin^2{x}$ in the equation and simplify to get a standard quadratic equation in $\sin{x}$.

Step 3: Solve the quadratic equation

Let $y = \sin{x}$. Solve the quadratic equation for $y$.

Step 4: Find the values of $x$

For each solution $y$, find all $x$ in the interval $0 \leq x \leq 2\pi$ such that $\sin{x} = y$.

Step 5: State the final solutions

List all values of $x$ that satisfy the original equation in the given interval.

Let's begin with Step 1 and Step 2.

Substitute $\cos^2{x} = 1 - \sin^2{x}$ into the equation:

$2(1 - \sin^2{x}) + 5\sin{x} = 4$

Now, let's expand and simplify:

$2 - 2\sin^2{x} + 5\sin{x} = 4$

$-2\sin^2{x} + 5\sin{x} + 2 - 4 = 0$

$-2\sin^2{x} + 5\sin{x} - 2 = 0$

Multiply both sides by $-1$ to make the quadratic standard:

$2\sin^2{x} - 5\sin{x} + 2 = 0$

Let $y = \sin{x}$:

$$2y^2 - 5y + 2 = 0$$

Now, let's solve this quadratic equation for $y$. Amanita, the solutions to the quadratic equation are:

$y = \frac{1}{2} \quad \text{and} \quad y = 2$

Recall that $y = \sin{x}$. However, the sine function only takes values between $-1$ and $1$. So, $\sin{x} = 2$ is not possible.

Now, let's focus on the valid solution:

$$\sin{x} = \frac{1}{2}$$

Next, let's find all values of $x$ in the interval $0 \leq x \leq 2\pi$ such that $\sin{x} = \frac{1}{2}$.

The general solution for $\sin{x} = \frac{1}{2}$ is:

$x = \arcsin{\left(\frac{1}{2}\right)} \quad \text{or} \quad x = \pi - \arcsin{\left(\frac{1}{2}\right)}$

Let's calculate these values. Amanita, let's clarify the values:

The value of $\arcsin{\left(\frac{1}{2}\right)}$ is $\frac{\pi}{6}$.

So, the two solutions for $x$ in the interval $0 \leq x \leq 2\pi$ are:

$$x_1 = \frac{\pi}{6}$$

$x_2 = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Therefore, the solutions to the equation $2\cos^2{x} + 5\sin{x} = 4$ for $0 \leq x \leq 2\pi$ are:

$x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}$