Find all solutions (in radians) of the equation 6 cos2 𝜃 + 7 sin 𝜃 − 8 = 0.

We are tasked with finding all solutions to the equation \( 6 \cos^2 \theta + 7 \sin \theta - 8 = 0 \), where \( \theta \) is in radians.

### Step 1: Use a trigonometric identity

Since we have both \( \cos^2 \theta \) and \( \sin \theta \) in the equation, we can use the Pythagorean identity:
\[
\cos^2 \theta = 1 - \sin^2 \theta
\]
Substitute this identity into the given equation:

\[
6(1 - \sin^2 \theta) + 7 \sin \theta - 8 = 0
\]

### Step 2: Simplify the equation

Distribute the 6:
\[
6 - 6 \sin^2 \theta + 7 \sin \theta - 8 = 0
\]
Combine like terms:
\[
-6 \sin^2 \theta + 7 \sin \theta - 2 = 0
\]

### Step 3: Solve the quadratic equation

This is a quadratic equation in \( \sin \theta \). Let \( x = \sin \theta \). We now have:
\[
-6x^2 + 7x - 2 = 0
\]
Multiply through by \(-1\) to simplify:
\[
6x^2 - 7x + 2 = 0
\]

### Step 4: Apply the quadratic formula

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For \( 6x^2 - 7x + 2 = 0 \), we have \( a = 6 \), \( b = -7 \), and \( c = 2 \). Plug these into the quadratic formula:
\[
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(2)}}{2(6)}
\]
\[
x = \frac{7 \pm \sqrt{49 - 48}}{12}
\]
\[
x = \frac{7 \pm \sqrt{1}}{12}
\]
\[
x = \frac{7 \pm 1}{12}
\]

### Step 5: Find the two solutions for \( x \)

Now, solving for the two possible values of \( x \):
1. \( x = \frac{7 + 1}{12} = \frac{8}{12} = \frac{2}{3} \)
2. \( x = \frac{7 - 1}{12} = \frac{6}{12} = \frac{1}{2} \)

### Step 6: Solve for \( \theta \)

We now solve \( \sin \theta = \frac{2}{3} \) and \( \sin \theta = \frac{1}{2} \).

#### Case 1: \( \sin \theta = \frac{2}{3} \)
To find \( \theta \), we take the inverse sine (arcsin):
\[
\theta_1 = \arcsin\left(\frac{2}{3}\right)
\]
Using the unit circle, the second solution occurs at:
\[
\theta_2 = \pi - \arcsin\left(\frac{2}{3}\right)
\]

#### Case 2: \( \sin \theta = \frac{1}{2} \)
For this, the standard values give:
\[
\theta_3 = \frac{\pi}{6}
\]
And the second solution in this case is:
\[
\theta_4 = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
\]

### Step 7: General solutions

Since sine is periodic with period \( 2\pi \), the general solutions are:

- For \( \sin \theta = \frac{2}{3} \):
  \[
  \theta = \arcsin\left(\frac{2}{3}\right) + 2n\pi \quad \text{or} \quad \theta = \pi - \arcsin\left(\frac{2}{3}\right) + 2n\pi
  \]
  where \( n \) is any integer.
  
- For \( \sin \theta = \frac{1}{2} \):
  \[
  \theta = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2n\pi
  \]

### Final Answer:

The solutions to the equation \( 6 \cos^2 \theta + 7 \sin \theta - 8 = 0 \) are:
\[
\theta = \arcsin\left(\frac{2}{3}\right) + 2n\pi, \quad \theta = \pi - \arcsin\left(\frac{2}{3}\right) + 2n\pi, \quad \theta = \frac{\pi}{6} + 2n\pi, \quad \theta = \frac{5\pi}{6} + 2n\pi
\]
where \( n \) is any integer.

Would you like more details or clarification on any part?

### Related questions:
1. How does the unit circle help in finding trigonometric solutions?
2. What is the significance of the general solution form in trigonometry?
3. How does the Pythagorean identity simplify trigonometric equations?
4. Can you explain the difference between arcsin and inverse trigonometric functions in general?
5. How do we determine when to use radians or degrees in solving trigonometric equations?

### Tip:
When solving trigonometric equations, always check if a substitution (like \( \cos^2 \theta = 1 - \sin^2 \theta \)) can simplify the equation before attempting more complex methods.

Find all solutions (in radians) of the equation 6 cos^2 θ + 7 sin θ − 8 = 0.

Discover how to solve the trigonometric equation 6 cos^2 θ + 7 sin θ − 8 = 0 step by step. This solution covers essential concepts such as the Pythagorean identity and quadratic formula, suitable for high school students in Grades 10-12.

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