If $\sin x+\cos x=\sqrt{2}\sin x$, then the value of $\sin x-\cos x$ is:

The correct answer is $\sqrt{2}\cos x$.

Given:

$\sin x+\cos x=\sqrt{2}\sin x$

Step 1: Square both sides of the given equation.

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

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

Step 2: Apply the Pythagorean identity $\sin ²x+\cos ²x=1$ on the left side.

$1+2\sin x\cos x=2\sin ²x$

Rearranging to isolate the product term:

$2\sin x\cos x=2\sin ²x-1$

Step 3: Now consider the expression we want to find.
Let's square $\sin x-\cos x$:

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

Again applying $\sin ²x+\cos ²x=1$:

$(\sin x-\cos x)²=1-2\sin x\cos x$

Step 4: Substitute the value of $2\sin x\cos x$ from Step 2.

$(\sin x-\cos x)²=1-(2\sin ²x-1)$

$(\sin x-\cos x)²=1-2\sin ²x+1$

$(\sin x-\cos x)²=2-2\sin ²x$

Step 5: Factor the right-hand side.

$(\sin x-\cos x)²=2(1-\sin ²x)$

Using the identity $1-\sin ²x=\cos ²x$:

$(\sin x-\cos x)²=2\cos ²x$

Step 6: Take the square root of both sides.

$\sin x-\cos x=\sqrt{2}\cos x$

Therefore, the value of $\sin x-\cos x$ is $\sqrt{2}\cos x$.