How to proof ,
$\displaystyle
(\sin x + \cos x)\sin 2x = \sqrt 2
$
thanks
$\displaystyle (\sin x + \cos x)\sin 2x = \sqrt 2$ cannot be proved because it is not true. I think that what you meant to ask is how to solve that equation for $\displaystyle x$
Try squaring both sides and expanding:
$\displaystyle (\sin x + \cos x)^2 \sin^2 2x = 2$
$\displaystyle (\sin^2 x + 2 \cos x \sin x + \cos^2 x) \sin^2 2x = 2$
$\displaystyle (1+ 2 \cos x \sin x) \sin^2 2x = 2$
$\displaystyle (1+ \sin 2x) \sin^2 2x = 2$
Letting $\displaystyle y = \sin 2x$, you have
$\displaystyle (1+ \sin 2x) \sin^2 2x = 2 \longrightarrow (1+y)y^2 = 2$
Solve for y by inspection. I see one obvious value of y such that $\displaystyle (1+y)y^2 = 2$
Then find when $\displaystyle \sin 2x$ equals that value of y and you're done.
Note: there is more than one solution. In fact, there are infinitely many solutions to this problem because sine is a periodic function.