Question

Решите пожалуйста. 1+ctg2x=1/cos(3п/2-2x)

Answer 1

$1+ctg2x= \frac{1}{\cos( \frac{3 \pi }{2}-2x) } \\ \\ 1+ctg2x= -\frac{1}{\sin2x} |\cdot \sin2x \\ \sin 2x+\cos 2x=-1$
Формула
$a \sin x\pm b\cos x= \sqrt{a^2+b^2} \sin (x\pm \arcsin \frac{b}{\sqrt{a^2+b^2}} )$

$\sqrt{a^2+b^2}= \sqrt{1^2+1^2} =\sqrt{2}$
$\arcsin \frac{b}{\sqrt{a^2+b^2}} =\arcsin \frac{1}{ \sqrt{2} } = \frac{\pi}{4}$

$\sqrt{2}\sin (2x+ \frac{\pi}{4} )=- \\ \sin(2x+ \frac{\pi}{4} )=- \frac{1}{\sqrt{2}} \\ 2x+\frac{\pi}{4}=(-1)^{k+1}\cdot \frac{\pi}{4}+ \pi k,k \in Z \\ 2x=(-1)^{k+1}\cdot \frac{\pi}{4}-\frac{\pi}{4}+\pi k,k \in Z \\ x=(-1)^{k+1}\cdot \frac{\pi}{8}-\frac{\pi}{8}+\frac{\pi k}{2}, k \in Z$