Question

Решите уравнение sin X - cos X = корень из 2 если X = [0; pi]

Answer 1

$\sin(x) - \cos(x) = \sqrt{2} \: \: \: | \times \frac{ \sqrt{2} }{2} \\ \frac{ \sqrt{2} }{2} \sin(x) - \frac{ \sqrt{2} }{2} \cos(x) = \frac{ \sqrt{2} \times \sqrt{2} }{2} \\ \cos( \frac{\pi}{4} ) \sin(x) - \sin( \frac{\pi}{4} ) \cos(x) = 1 \\ \sin(x - \frac{\pi}{4} ) = 1 \\ x - \frac{\pi}{4} = \frac{\pi}{2} + 2\pi \: n \\ x = \frac{3\pi}{4} + 2\pi \: n$

n принадлежит Z.

на [0;П]

$x = \frac{3\pi}{4} \\$

Answer 2

$sinx-cosx=\sqrt2\\\\sinx-sin\Big(\dfrac{\pi}{2}-x\Big)=\sqrt2\\\\2\cdot sin\Big(x-\dfrac{\pi}{4}\Big)\cdot cos\dfrac{\pi}{4}=\sqrt2\\\\2\cdot sin\Big(x-\dfrac{\pi}{4}\Big)\cdot \dfrac{\sqrt2}{2}=\sqrt2\\\\sin\Big(x-\dfrac{\pi}{4}\Big)=1\\\\x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\\\\x=\dfrac{3\pi }{4}+2\pi n\ ,\ n\in Z\\\\x\in [\, 0\, ;\, \pi \ ]\ \ \Rightarrow \ \ x=\dfrac{3\pi }{4}$