Solve `4 cos^(2)x+6 sin^(2)x=5`.

Explanation

To solve the equation \( 4 \cos^2 x + 6 \sin^2 x = 5 \), we can follow these steps:

### Step 1: Use the Pythagorean Identity
We know from the Pythagorean identity that:
\[ \sin^2 x + \cos^2 x = 1 \]
We can express \( \cos^2 x \) in terms of \( \sin^2 x \):
\[ \cos^2 x = 1 - \sin^2 x \]

### Step 2: Substitute into the Equation
Substituting \( \cos^2 x \) into the original equation gives:
\[ 4(1 - \sin^2 x) + 6 \sin^2 x = 5 \]

### Step 3: Simplify the Equation
Now, distribute the 4:
\[ 4 - 4 \sin^2 x + 6 \sin^2 x = 5 \]
Combine like terms:
\[ 4 + 2 \sin^2 x = 5 \]

### Step 4: Isolate the \( \sin^2 x \) Term
Subtract 4 from both sides:
\[ 2 \sin^2 x = 5 - 4 \]
This simplifies to:
\[ 2 \sin^2 x = 1 \]

### Step 5: Solve for \( \sin^2 x \)
Divide both sides by 2:
\[ \sin^2 x = \frac{1}{2} \]

### Step 6: Take the Square Root
Taking the square root of both sides gives:
\[ \sin x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

### Step 7: Find the General Solutions
The angles for which \( \sin x = \frac{\sqrt{2}}{2} \) are:
\[ x = \frac{\pi}{4} + 2n\pi \quad \text{and} \quad x = \frac{3\pi}{4} + 2n\pi \]
The angles for which \( \sin x = -\frac{\sqrt{2}}{2} \) are:
\[ x = -\frac{\pi}{4} + 2n\pi \quad \text{and} \quad x = -\frac{3\pi}{4} + 2n\pi \]

### Final General Solution
Thus, the general solutions can be summarized as:
\[ x = \frac{\pi}{4} + 2n\pi, \quad x = \frac{3\pi}{4} + 2n\pi, \quad x = -\frac{\pi}{4} + 2n\pi, \quad x = -\frac{3\pi}{4} + 2n\pi \]
where \( n \) is any integer.