I need to bring this equation \sin x + \cos x - \sin x \cos x - m = 0

to this form: 2 \cos^2 \( x - \frac{\pi}{4} \) - 2 \sqrt{2} \cos \( x - \frac{\pi}{4} \) +2m -1 = 000

How do I get there?

Start on the second equation by using the identity cos(a-b) = cos(a)cos(b+sin(a)sin(b), where a = x and b = pi/4. The only other identity you'll need is cos^2(x) + sin^2(x) = 1. It works right out.

Thank you, Ebaines! I did work out, indeed! :)

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

\Rightarrow \sin x \cos x = \frac{(\sin x + \cos x)^2 -1 }{2}

\cos \(x - \frac{\pi}{4}\) = \frac{\sqrt{2}}{2} (\cos x + \sin x)

The first equation can be written:
(\sin x + \cos x)^2 - 2(\sin x + \cos x) -2m -1 = 0

\[ \sqrt{2} \cos \(x - \frac{\pi}{4} \) \]^2 - 2 \sqrt{2} \cos \( x - \frac{\pi}{4} \) - 2m -1 = 0

... and here is the final form:
2 \cos^2 \( x- \frac{\pi}{4} \) - 2 \sqrt{2} \cos \( x- \frac{\pi}{4} \) -2m -1 =0