Применение основных тригонометрических формул для решения уравнений. Задача 24

ОТВЕТ: \(\dfrac{\pi }{2} + \pi k;\quad \,-\dfrac{\pi }{4} + 2\pi k;\quad -\dfrac{{3\pi }}{4} + 2\pi k;\quad k \in Z.\)

\(\sin 2x + \sqrt 2 \cos x = 0\)

Воспользуемся формулой синуса двойного угла:

\(\sin 2\alpha  = 2\sin \alpha \cos \alpha .\)

\(\sin 2x + \sqrt 2 \cos x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\sin x\cos x + \sqrt 2 \cos x = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\cos x\left( {2\sin x + \sqrt 2 } \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2\sin x + \sqrt 2  = 0}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos x = 0,\,\,\,\,\,\,\,}\\{\sin x = -\dfrac{{\sqrt 2 }}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{2} + \pi k,\,\,\,\,\,\,\,\,\,\,}\\{x = -\dfrac{\pi }{4} + 2\pi k,\,\,\,}\\{x = -\dfrac{{3\pi }}{4} + 2\pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\dfrac{\pi }{2} + \pi k;\quad \,-d\frac{\pi }{4} + 2\pi k;\quad -\dfrac{{3\pi }}{4} + 2\pi k;\quad k \in Z.\)