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The given equation is:

\[
20 \sin^2 x = 42 \cos x - 6
\]

We are tasked with finding the two values of \(x\) in the interval \([0, 2\pi]\) that satisfy this equation.

### Step 1: Simplify the equation
We begin by expressing \(\sin^2 x\) in terms of \(\cos x\). Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute into the equation:

\[
20 (1 - \cos^2 x) = 42 \cos x - 6
\]

Simplify this:

\[
20 - 20 \cos^2 x = 42 \cos x - 6
\]

Rearrange all terms to one side:

\[
20 \cos^2 x + 42 \cos x - 26 = 0
\]

### Step 2: Solve the quadratic equation
This is now a quadratic equation in \(\cos x\):

\[
20 \cos^2 x + 42 \cos x - 26 = 0
\]

Use the quadratic formula:

\[
\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where:
- \(a = 20\)
- \(b = 42\)
- \(c = -26\)

Substitute these values:

\[
\cos x = \frac{-42 \pm \sqrt{42^2 - 4(20)(-26)}}{2(20)}
\]
\[
\cos x = \frac{-42 \pm \sqrt{1764 + 2080}}{40}
\]
\[
\cos x = \frac{-42 \pm \sqrt{3844}}{40}
\]
\[
\cos x = \frac{-42 \pm 62}{40}
\]

### Step 3: Calculate the two solutions
Now, solve for the two values of \(\cos x\):

1. \(\cos x = \frac{-42 + 62}{40} = \frac{20}{40} = \frac{1}{2}\)
2. \(\cos x = \frac{-42 - 62}{40} = \frac{-104}{40} = -\frac{26}{10} = -2.6\)

Since \(\cos x\) must be in the range \([-1, 1]\), the second solution (\(\cos x = -2.6\)) is not valid. Therefore, we are left with:

\[
\cos x = \frac{1}{2}
\]

### Step 4: Find the values of \(x\)
We now find \(x\) such that \(\cos x = \frac{1}{2}\) in the interval \([0, 2\pi]\). The solutions are:

\[
x = \frac{\pi}{3}, \quad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}
\]

### Final Answer:
- The smaller value of \(x\) is \(\frac{\pi}{3}\).
- The larger value of \(x\) is \(\frac{5\pi}{3}\).

Would you like further details or clarifications on any step?

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Here are five related questions:
1. How do you solve other trigonometric equations like \( \sin^2 x + \cos x = 1 \)?
2. What are the main trigonometric identities used to simplify complex equations?
3. How does the quadratic formula apply to trigonometric equations?
4. What other methods can be used to solve trigonometric equations apart from algebraic methods?
5. How can the unit circle help in finding solutions to trigonometric equations?

**Tip:** Always check the domain of the trigonometric function values when solving equations, as it helps eliminate invalid solutions.

Consider the equation 20sin^2x = 42cosx - 6. There are two values of x in the interval [0, 2π] that satisfy this equation. Find those values. Write the exact value, but not the approximate value. Write Pi, pi, or π if your answer contains π. Write the smaller and bigger value of x in the interval [0, 2π] that satisfy the equation.

Solve the trigonometric equation 20sin²x = 42cosx - 6 in the interval [0, 2π]. Learn step-by-step how to use trigonometric identities and the quadratic formula to find the exact values of x. Suitable for students in Grades 10-12.

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