Question

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Answer 1

Ответ:

$2\sin 2x=\sin x-\sqrt3\cos x\\\\2\sin 2x=2\cdot \frac12 \sin x-2*\frac{\sqrt3}2 \cos x\\\\2\sin 2x=2(\cos \frac\pi3\sin x- \sin \frac\pi3 \cos x)\\\\2\sin 2x=2\cos(x-\frac\pi3)\\\\\sin 2x-\cos(x-\frac\pi3)=0\\\\\\2\cos\frac{2x+(x-\frac\pi3)}2\cdot \sin\frac{2x-(x-\frac\pi3)}2=0\\\\\\cos(\frac32x-\frac\pi6) \cdot \sin(\frac12x+\frac\pi6)=0\\\\\cos(\frac32x-\frac\pi6)=0\,\,\,\,\,\,\,\,\,\,\,\, \sin(\frac12x+\frac\pi6)=0\\\\\frac32x-\frac\pi6=\frac\pi2+\frac{\pi k}2\,\,\,\,\, \frac12x+\frac\pi6=\pi n$

$\frac32x=\frac{2\pi}3+\frac{\pi k}2\,\,\,\,\, \frac12x=-\frac\pi6+\pi n\\\\\\x=\frac{4\pi}9+\frac{\pi k }3\,\,\,\,\,\, x=-\frac\pi3+2\pi n$

Пошаговое объяснение: