2. $$ 2 \sin ^{2} x=2+\cos x $$

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To solve the trigonometric equation:

$$
2 \sin^2 x = 2 + \cos x
$$

follow these steps:

### **Step 1: Use a Pythagorean Identity**

Recall the Pythagorean identity:

$$
\sin^2 x = 1 - \cos^2 x
$$

Substitute this into the original equation:

$$
2(1 - \cos^2 x) = 2 + \cos x
$$

### **Step 2: Simplify the Equation**

Expand and simplify:

$$
2 - 2\cos^2 x = 2 + \cos x
$$

Subtract $$2$$ from both sides:

$$
-2\cos^2 x = \cos x
$$

Bring all terms to one side to form a quadratic equation:

$$
2\cos^2 x + \cos x = 0
$$

### **Step 3: Factor the Equation**

Factor out $$\cos x$$:

$$
\cos x (2\cos x + 1) = 0
$$

This gives two possibilities:

1. $$\cos x = 0$$
2. $$2\cos x + 1 = 0$$ → $$\cos x = -\frac{1}{2}$$

### **Step 4: Solve for $$x$$**

#### **Case 1: $$\cos x = 0$$**

The cosine of an angle is zero at:

$$
x = \frac{\pi}{2} + k\pi \quad \text{where } k \text{ is any integer}
$$

#### **Case 2: $$\cos x = -\frac{1}{2}$$**

The cosine of an angle is $$-\frac{1}{2}$$ at:

$$
x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi \quad \text{where } k \text{ is any integer}
$$

### **Step 5: Combine the Solutions**

All solutions to the equation are:

$$
x = \frac{\pi}{2} + k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2k\pi \quad \text{where } k \text{ is any integer}
$$

### **Final Answer**

All real numbers congruent to π⁄2 modulo π or to 2π⁄3 or 4π⁄3 modulo 2π. In other words, the solutions are

x = π/2 + kπ

or

x = 2π/3 + 2kπ

or

x = 4π/3 + 2kπ

for any integer k.
The solutions are: - $$ x = \frac{\pi}{2} + k\pi $$ - $$ x = \frac{2\pi}{3} + 2k\pi $$ - $$ x = \frac{4\pi}{3} + 2k\pi $$ where $$ k $$ is any integer.

2. \( 2 \sin ^{2} x=2+\cos x \)

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