8cos2x – 6cos x = -1 solve for when 0°<=x<=360°

To solve the equation \( 8\cos(2x) - 6\cos(x) = -1 \) for \( x \) in the interval \( 0^\circ \leq x \leq 360^\circ \), let's proceed step by step.

### Step 1: Simplify the equation
First, let's rewrite the equation:
\[
8\cos(2x) - 6\cos(x) = -1
\]

Add 1 to both sides to simplify:
\[
8\cos(2x) - 6\cos(x) + 1 = 0
\]

### Step 2: Use the double-angle identity
Recall the double-angle identity for cosine:
\[
\cos(2x) = 2\cos^2(x) - 1
\]

Substitute this into the equation:
\[
8(2\cos^2(x) - 1) - 6\cos(x) + 1 = 0
\]

### Step 3: Expand and simplify
Expanding the equation gives:
\[
16\cos^2(x) - 8 - 6\cos(x) + 1 = 0
\]

Simplify further:
\[
16\cos^2(x) - 6\cos(x) - 7 = 0
\]

### Step 4: Solve the quadratic equation
This is a quadratic equation in terms of \(\cos(x)\):
\[
16\cos^2(x) - 6\cos(x) - 7 = 0
\]

We can solve this quadratic equation using the quadratic formula:
\[
\cos(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 16 \), \( b = -6 \), and \( c = -7 \).

Substituting the values:
\[
\cos(x) = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(16)(-7)}}{2(16)}
\]

\[
\cos(x) = \frac{6 \pm \sqrt{36 + 448}}{32}
\]

\[
\cos(x) = \frac{6 \pm \sqrt{484}}{32}
\]

\[
\cos(x) = \frac{6 \pm 22}{32}
\]

This gives two possible solutions:
\[
\cos(x) = \frac{28}{32} = \frac{7}{8} \quad \text{or} \quad \cos(x) = \frac{-16}{32} = -\frac{1}{2}
\]

### Step 5: Find the corresponding angles
#### Case 1: \( \cos(x) = \frac{7}{8} \)
The value \( \cos(x) = \frac{7}{8} \) corresponds to the angles:
\[
x \approx \cos^{-1}\left(\frac{7}{8}\right)
\]
Using a calculator:
\[
x \approx 28.96^\circ \quad \text{and} \quad x \approx 360^\circ - 28.96^\circ = 331.04^\circ
\]

#### Case 2: \( \cos(x) = -\frac{1}{2} \)
The value \( \cos(x) = -\frac{1}{2} \) corresponds to the angles:
\[
x = 120^\circ \quad \text{and} \quad x = 240^\circ
\]

### Final Answer:
The solutions to the equation \( 8\cos(2x) - 6\cos(x) = -1 \) within the interval \( 0^\circ \leq x \leq 360^\circ \) are:
\[
x \approx 28.96^\circ, \quad x \approx 331.04^\circ, \quad x = 120^\circ, \quad x = 240^\circ
\]

Would you like more details on any of these steps, or have any other questions?

Here are some related questions you might explore:
1. How do you derive the quadratic formula?
2. What are the properties of the cosine function?
3. How do you solve trigonometric equations involving double angles?
4. What are the applications of the cosine double-angle identity?
5. How do you determine the angles corresponding to specific cosine values?

**Tip:** When solving trigonometric equations, always ensure your calculator is in the correct mode (degrees or radians) corresponding to the problem's context.

Learn how to solve the trigonometric equation 8cos(2x) - 6cos(x) = -1 within the interval 0° to 360°. Step-by-step solution using the double-angle identity for cosine and quadratic formula. Suitable for students in Grades 10-12.

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