when 0°<= x <= 360°, 6 cos² x = cos x + 5, how many root in that formula?

To solve the equation \(6 \cos^2 x = \cos x + 5\) for \(0^\circ \leq x \leq 360^\circ\), we first rewrite it as a quadratic equation in terms of \(\cos x\).

### Step 1: Rewrite the equation
\[
6 \cos^2 x - \cos x - 5 = 0
\]
Let \(y = \cos x\). The equation becomes:
\[
6y^2 - y - 5 = 0
\]

### Step 2: Solve the quadratic equation
We solve \(6y^2 - y - 5 = 0\) using the quadratic formula:
\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-5)}}{2(6)}
\]
\[
y = \frac{1 \pm \sqrt{1 + 120}}{12}
\]
\[
y = \frac{1 \pm \sqrt{121}}{12}
\]
\[
y = \frac{1 \pm 11}{12}
\]
Thus, we have two solutions for \(y\):
\[
y = \frac{1 + 11}{12} = 1 \quad \text{and} \quad y = \frac{1 - 11}{12} = -\frac{5}{6}
\]

### Step 3: Solve for \(x\)
Now we solve for \(x\) in each case:

1. **For \(y = 1\):**
   \[
   \cos x = 1
   \]
   This happens when \(x = 0^\circ\).

2. **For \(y = -\frac{5}{6}\):**
   \[
   \cos x = -\frac{5}{6}
   \]
   The cosine function is negative in the second and third quadrants. Using \(\cos^{-1} \left(-\frac{5}{6}\right)\), we get:

   \[
   x = \cos^{-1}\left(-\frac{5}{6}\right) \approx 146.42^\circ \quad (\text{in the second quadrant})
   \]
   \[
   x = 360^\circ - 146.42^\circ \approx 213.58^\circ \quad (\text{in the third quadrant})
   \]

### Step 4: Conclusion
The solutions for \(x\) are:
\[
x = 0^\circ, 146.42^\circ, 213.58^\circ
\]
Thus, there are **3 roots** in the interval \(0^\circ \leq x \leq 360^\circ\).

Would you like more details or have any other questions?

### Related Questions:
1. How do we apply the quadratic formula in general?
2. What are the properties of the cosine function for specific intervals?
3. How do you find solutions to trigonometric equations in different quadrants?
4. What is the significance of the inverse cosine function?
5. How can we check the validity of roots for trigonometric equations?

**Tip:** When solving quadratic equations involving trigonometric functions, always check which quadrants the solutions fall in to ensure correctness.

When 0° <= x <= 360°, 6 cos² x = cos x + 5, how many roots in that formula?

Explore the step-by-step solution to solving the trigonometric equation 6 cos² x = cos x + 5 within the interval 0° <= x <= 360°. Learn how to apply the quadratic formula, solve for cosine, and find roots using inverse trigonometric functions. Ideal for students in Grades 10-12.

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