The equation provided in the image is:

$\sqrt{1 + \sin(2x)} = \sin(x) + \cos(x)$

To solve this equation, let's follow these steps:

Step 1: Square Both Sides

Square both sides to eliminate the square root:

$\left(\sqrt{1 + \sin(2x)}\right)^2 = (\sin(x) + \cos(x))^2$

This simplifies to:

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

Step 2: Simplify the Right Side

Use the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$:

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

Notice that $\sin(2x) = 2\sin(x)\cos(x)$, so the equation becomes:

$1 + \sin(2x) = 1 + \sin(2x)$

Step 3: Analyze the Result

Both sides of the equation are identical, which means the equation holds true for all $x$. Therefore, the solution set includes all real numbers:

$\boxed{x \in \mathbb{R}}$

This means the equation is an identity, and it is valid for any real value of $x$.

Would you like more details on any part of this solution or have any further questions?

Follow-Up Questions:

1. What is the Pythagorean identity and how is it used in trigonometric equations?
2. Can you provide an example where squaring both sides of an equation introduces extraneous solutions?
3. What is the double-angle identity for sine, and how is it derived?
4. How would you verify if the equation holds for specific values of $x$?
5. What are some common trigonometric identities that are frequently used in solving equations?

Tip:

Always check for extraneous solutions when squaring both sides of an equation, as this can sometimes introduce invalid solutions.