Question

Помогите решить Sin (pi/6 -x)*sinx=sin2x

Answer 1

Sin (pi/6 - x)*sin x = sin 2x = 2sin x*cos x
sin x*(sin(pi/6 - x) - 2cos x) = 0
1) sin x = 0; x1 = pi*k
2) sin(pi/6 - x) - 2cos x = 0
sin(pi/6)*cos x - cos(pi/6)*sin x - 2cos x = 0
1/2*cos x - √3/2*sin x - 2cos x = 0
-√3/2*sin x - 3/2*cos x = 0
-√3*sin x = 3cos x
tg x = -3/√3 = -√3; x2 = -pi/3 + pi*k
Ответ: x1 = pi*k; x2 = -pi/3 + pi*k

Answer 2

$sin( \frac{\pi }{6}-x)\cdot sinx=sin2x\\\\sin( \frac{\pi }{6}-x)\cdot sinx-2sinx\cdot cosx=0\\\\sinx\cdot (sin( \frac{\pi}{6}-x)-2cosx)=0\\\\a)\; \; sinx=0\; \; \to \; \; \; x=\pi n,\; n\in Z\\\\b)\; \; sin \frac{\pi }{6}\cdt cosx-cos\frac{\pi }{6}\cdot sinx-2cosx=0\\\\\frac{1}{2}\cdot cosx-\frac{\sqrt3}{2}\cdot sinx-2cosx=0\\\\- \frac{\sqrt3}{2}\cdot sinx-\frac{3}{2}\cdot cosx=0\; |\cdot (-2)\\\\\sqrt3\cdot (sinx+\sqrt3\cdot cosx)=0\\\\sinx+\sqrt3\cdot cosx=0\; |:2\\\\ \frac{1}{2}\cdot sinx+\frac{\sqrt3}{2}\cdot cosx=0$

$cos \frac{\pi }{3}\cdot sinx+sin\frac{\pi }{3}\cdot cosx=0\\\\sin( \frac{\pi }{3}+x)=0\\\\ \frac{\pi }{3}+x=\pi k\; ,\; k\in Z\\\\ x=- \frac{\pi }{3}+\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=\pi n\; ,\; \; x=- \frac{\pi }{3}+\pi k\; ,\; \; n,k\in Z\; .$