Question

Очень нудна помощь в решении

Answer 1

Решить систему тригонометрических уравнений.

$\displaystyle \left \{ {{\cos x\cos y=\sin^2y} \atop {\sin x\sin y=\cos^2y}} \right. ~~~\Big|~+\\\\ \left \{ {{\cos x\cos y+\sin x\sin y=\sin^2y+\cos^2y} \atop {\sin x\sin y=\cos^2y}} \right.\\\\\\ \left \{ {{\cos x\cos y+\sin x\sin y=1} \atop {\sin x\sin y=\cos^2y}} \right.~~~~ \left \{ {{\cos (x- y)=1} \atop {\sin x\sin y=\cos^2y}} \right.\\\\\\ \left \{ {{x- y=2\pi n} \atop {\sin x\sin y=\cos^2y}} \right.~~~~ \left \{ {{x= y+2\pi n} \atop {\sin (y+ 2\pi n)\sin y=\cos^2y}} \right.$

$\displaystyle \left \{ {{x= y+2\pi n} \atop {\sin y \sin y=1-\sin^2y}} \right.~~~~ \left \{ {{x= y+2\pi n} \atop {2\sin^2 y=1}} \right.\\\\\\ \left \{ {{x= y+2\pi n} \atop {\sin^2 y=\dfrac 12}} \right.~~~~ \left \{ {{x= y+2\pi n} \atop {\sin y=\pm \dfrac 1{\sqrt2}}} \right.~~~\Leftrightarrow~~~ \left \{ {{x= y+2\pi n} \atop {\sin y=\pm \dfrac {\sqrt2}2}} \right.$

$\boxed {y=\dfrac {\pi}4+\dfrac{\pi}2 k,~~~x=\dfrac {\pi}4+\dfrac{\pi}2 k+2\pi n;~~~k,n\in Z }$

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Использованы формулы

$\sin^2 y + \cos^2 y = 1\\ \cos x \cos y + \sin x \sin y = \cos (x-y)$