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Let's solve the equation step-by-step. The equation given is:

\[
20\sin^2(x) - 7\sin(x) - 3 = 0
\]

### Step 1: Substitution
We can use the substitution \(u = \sin(x)\) to convert the equation into a quadratic form:

\[
20u^2 - 7u - 3 = 0
\]

### Step 2: Solve the quadratic equation
Using the **quadratic formula**:

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

where \(a = 20\), \(b = -7\), and \(c = -3\). Substituting these values:

\[
u = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(20)(-3)}}{2(20)}
\]

\[
u = \frac{7 \pm \sqrt{49 + 240}}{40} = \frac{7 \pm \sqrt{289}}{40}
\]

\[
u = \frac{7 \pm 17}{40}
\]

So the two solutions for \(u\) are:

\[
u = \frac{7 + 17}{40} = \frac{24}{40} = \frac{3}{5}, \quad u = \frac{7 - 17}{40} = \frac{-10}{40} = -\frac{1}{4}
\]

### Step 3: Back-substitute \(u = \sin(x)\)
Now, we solve for \(x\) in the interval \([0, 2\pi]\).

#### Case 1: \( \sin(x) = \frac{3}{5} \)
We look for the angles where \(\sin(x) = \frac{3}{5}\).

\[
x = \sin^{-1}\left(\frac{3}{5}\right)
\]

Using the inverse sine function, the principal solution is:

\[
x_1 = \sin^{-1}\left(\frac{3}{5}\right)
\]

In the interval \([0, 2\pi]\), the two solutions are:

\[
x_1 = \sin^{-1}\left(\frac{3}{5}\right), \quad x_2 = \pi - \sin^{-1}\left(\frac{3}{5}\right)
\]

#### Case 2: \( \sin(x) = -\frac{1}{4} \)
Now, we solve for the angles where \(\sin(x) = -\frac{1}{4}\).

\[
x = \sin^{-1}\left(-\frac{1}{4}\right)
\]

The principal solution is:

\[
x_3 = -\sin^{-1}\left(\frac{1}{4}\right)
\]

In the interval \([0, 2\pi]\), the two solutions are:

\[
x_3 = 2\pi + \sin^{-1}\left(-\frac{1}{4}\right), \quad x_4 = \pi - \sin^{-1}\left(-\frac{1}{4}\right)
\]

### Step 4: Final Solutions
Thus, the exact solutions for the equation \(20\sin^2(x) - 7\sin(x) - 3 = 0\) in the interval \([0, 2\pi]\) are:

\[
x_1 = \sin^{-1}\left(\frac{3}{5}\right), \quad x_2 = \pi - \sin^{-1}\left(\frac{3}{5}\right)
\]

\[
x_3 = 2\pi + \sin^{-1}\left(-\frac{1}{4}\right), \quad x_4 = \pi - \sin^{-1}\left(-\frac{1}{4}\right)
\]

Would you like further help or a numerical approximation of these solutions? 

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**5 Related Questions:**
1. How do you evaluate inverse trigonometric functions?
2. What are the properties of the sine function on the unit circle?
3. How do you solve quadratic equations systematically?
4. How can trigonometric identities help in solving similar equations?
5. How do we verify solutions for trigonometric equations?

**Tip:** For trigonometric equations, always double-check if solutions lie within the required interval.

Solve the equation on the interval [0, 2π). Give your answers as exact solutions. Write all numbers as integers or simplified fractions.

20sin²(x) - 7sin(x) - 3 = 0

Learn to solve the trigonometric equation 20sin²(x) - 7sin(x) - 3 = 0 on the interval [0, 2π) with exact solutions. This solution involves quadratic equations and inverse trigonometric functions, suitable for students in Grades 10-12.

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