For every real number find all the real solutions to equation ` sin x + cos(a+x)+ cos (a-x)=2 `

Answer

To solve the equation \( \sin x + \cos(a+x) + \cos(a-x) = 2 \), we will follow these steps:

### Step 1: Simplify the equation
We start with the given equation:
\[ \sin x + \cos(a+x) + \cos(a-x) = 2 \]

Using the cosine addition formula, we know that:
\[ \cos(a+x) + \cos(a-x) = 2 \cos a \cos x \]
Thus, we can rewrite the equation as:
\[ \sin x + 2 \cos a \cos x = 2 \]

### Step 2: Rearranging the equation
Now, we rearrange the equation:
\[ \sin x + 2 \cos a \cos x - 2 = 0 \]

### Step 3: Identify coefficients
We can identify this as a standard form of a trigonometric equation:
\[ \sin x + b \cos x = c \]
where \( b = 2 \cos a \) and \( c = 2 \).

### Step 4: Apply the condition for real solutions
For the equation \( \sin x + b \cos x = c \) to have real solutions, we need to satisfy the condition:
\[ \left| c \right| \leq \sqrt{a^2 + b^2} \]
In our case:
\[ \left| 2 \right| \leq \sqrt{1^2 + (2 \cos a)^2} \]
This simplifies to:
\[ 2 \leq \sqrt{1 + 4 \cos^2 a} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ 4 \leq 1 + 4 \cos^2 a \]
Subtracting 1 from both sides results in:
\[ 3 \leq 4 \cos^2 a \]
Dividing both sides by 4:
\[ \frac{3}{4} \leq \cos^2 a \]

### Step 6: Find the range for \( \cos a \)
Taking the square root gives:
\[ \sqrt{\frac{3}{4}} \leq | \cos a | \]
This implies:
\[ \cos a \geq \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos a \leq -\frac{\sqrt{3}}{2} \]

### Step 7: Determine the angles
The values of \( a \) for which \( \cos a \) meets these criteria are:
\[ a \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \cup \left[\frac{5\pi}{6}, \frac{7\pi}{6}\right] \]

### Final Solution
Thus, the values of \( a \) that satisfy the original equation are:
\[ a \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \cup \left[\frac{5\pi}{6}, \frac{7\pi}{6}\right] \]

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