We are tasked with solving the trigonometric equation:

\[
3\sin^2(x) - \sin(x) - 1 = 0
\]

### Step 1: Substitution
To simplify the equation, let's make a substitution:
\[
u = \sin(x)
\]
Thus, the equation becomes:
\[
3u^2 - u - 1 = 0
\]
This is a quadratic equation in \( u \).

### Step 2: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \( 3u^2 - u - 1 = 0 \), the coefficients are:
- \( a = 3 \)
- \( b = -1 \)
- \( c = -1 \)

Substituting these values into the quadratic formula:
\[
u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-1)}}{2(3)}
\]
\[
u = \frac{1 \pm \sqrt{1 + 12}}{6}
\]
\[
u = \frac{1 \pm \sqrt{13}}{6}
\]
Thus, the two solutions for \( u \) are:
\[
u_1 = \frac{1 + \sqrt{13}}{6} \quad \text{and} \quad u_2 = \frac{1 - \sqrt{13}}{6}
\]

### Step 3: Interpret the solutions
Since \( u = \sin(x) \), we know that the sine function must have values between -1 and 1. Therefore, we reject the solution \( u_1 = \frac{1 + \sqrt{13}}{6} \), because it is greater than 1.

Now we check the second solution:
\[
u_2 = \frac{1 - \sqrt{13}}{6} \approx -0.767
\]
This is within the valid range for \( \sin(x) \).

### Step 4: Solve for \( x \)
Now, we need to solve for \( x \) where:
\[
\sin(x) = -0.767
\]
The general solution for \( \sin(x) = -0.767 \) is:
\[
x = \arcsin(-0.767) + 2n\pi \quad \text{or} \quad x = \pi - \arcsin(-0.767) + 2n\pi
\]
for any integer \( n \).

Using a calculator:
\[
x \approx \arcsin(-0.767) \approx -0.872 \text{ radians}
\]
Thus, the two solutions are:
\[
x \approx -0.872 + 2n\pi \quad \text{or} \quad x \approx 3.141 - (-0.872) + 2n\pi = 4.013 + 2n\pi
\]

### Final Answer:
The general solutions are:
\[
x \approx -0.872 + 2n\pi \quad \text{or} \quad x \approx 4.013 + 2n\pi
\]
for any integer \( n \).

Would you like any details or have questions about the steps?

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Here are 5 related questions to extend your understanding:

1. How do you solve quadratic equations using factoring instead of the quadratic formula?
2. What is the geometric interpretation of \( \sin(x) \) on the unit circle?
3. Can the sine function have values greater than 1 or less than -1? Why or why not?
4. How would the solution change if the equation involved \( \cos(x) \) instead of \( \sin(x) \)?
5. How do you determine the period of a trigonometric function like \( \sin(x) \)?

**Tip:** Always check if your trigonometric solution is within the valid range for the sine or cosine function.

Explore how to solve the trigonometric quadratic equation 3sin^2(x) - sin(x) - 1 = 0 using the quadratic formula. This problem covers key concepts in trigonometry and quadratic equations, ideal for students in Grades 10-12.

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