Solve the following equation for all values of 0, is less than or equal to, theta, is less than, 2, pi0≤θ<2π.
4, sine, squared, theta, minus, 6, sine, theta, minus, 4, equals, 0
4sin 
2
 θ−6sinθ−4=

The equation given is:

\[
4\sin^2(\theta) - 6\sin(\theta) - 4 = 0
\]

Let's solve this step by step:

### Step 1: Let \( x = \sin(\theta) \)

This will transform the equation into a quadratic form:

\[
4x^2 - 6x - 4 = 0
\]

### Step 2: Solve the quadratic equation

Use the quadratic formula to solve for \( x \):

\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(4)(-4)}}{2(4)}
\]

Simplify the expression:

\[
x = \frac{6 \pm \sqrt{36 + 64}}{8}
\]
\[
x = \frac{6 \pm \sqrt{100}}{8}
\]
\[
x = \frac{6 \pm 10}{8}
\]

Thus, we have two possible solutions for \( x \):

\[
x_1 = \frac{6 + 10}{8} = \frac{16}{8} = 2
\]
\[
x_2 = \frac{6 - 10}{8} = \frac{-4}{8} = -\frac{1}{2}
\]

### Step 3: Analyze the solutions

- \( x_1 = 2 \) is not possible because \( \sin(\theta) \) can only take values between \(-1\) and \(1\), so we discard this solution.
- \( x_2 = -\frac{1}{2} \) is valid, so we solve for \( \theta \) where \( \sin(\theta) = -\frac{1}{2} \).

### Step 4: Find \( \theta \) for \( \sin(\theta) = -\frac{1}{2} \)

The general solution for \( \sin(\theta) = -\frac{1}{2} \) occurs at:

\[
\theta = \frac{7\pi}{6}, \frac{11\pi}{6}
\]

These are the two angles where \( \sin(\theta) = -\frac{1}{2} \) within the interval \( 0 \leq \theta < 2\pi \).

### Final Answer:

The solutions for \( \theta \) are:

\[
\theta = \frac{7\pi}{6}, \frac{11\pi}{6}
\]

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Do you want more details or have any questions?

Here are 5 related questions:
1. How can you derive the general solutions for trigonometric equations?
2. What is the significance of the unit circle in solving trigonometric equations?
3. How does the range of sine and cosine functions affect their solutions?
4. Can you explain the quadratic formula in more detail?
5. How would this solution change if the equation was \( \cos(\theta) \) instead of \( \sin(\theta) \)?

**Tip**: Always check the domain restrictions of trigonometric functions when solving equations!

Solve the following equation for all values of 0 ≤ θ < 2π: 4sin^2(θ) - 6sin(θ) - 4 = 0.

This page provides a step-by-step solution for the trigonometric equation 4sin^2(θ) - 6sin(θ) - 4 = 0, covering key concepts in trigonometry and quadratic equations. Learn how to solve for θ within the interval 0 ≤ θ < 2π. Suitable for students in Grades 10-12.

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