Пояснення:
[tex]\displaystyle2*(cosx)^2+\sqrt{3}*sin(2x)=0 \\\\2*cos^2x+\sqrt{3}*2*sinx*cosx=0\ |:2\\\\cosx*(cosx+\sqrt{3} *sinx)=0\\\\cosx=0\\\\x_1=\frac{\pi }{2}+\pi n,\ \ \ n\in Z .\\\\cosx+\sqrt{3}*sinx=0\ |:2\\\\cosx*\frac{1}{2} +sinx*\frac{\sqrt{3} }{2} =0\\\\cosx*cos\frac{\pi }{3} +sinx*sin\frac{\pi }{3} =0\\\\cos(x-\frac{\pi }{3} )=0\\\\x-\frac{\pi }{3} =\frac{\pi }{2}+\pi n\\\\x_2=\frac{\pi }{2}+\frac{\pi }{3} +\pi n=\frac{5\pi }{6} +\pi n,\ \ \ n\in Z.[/tex]
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
Пояснення:
[tex]\displaystyle2*(cosx)^2+\sqrt{3}*sin(2x)=0 \\\\2*cos^2x+\sqrt{3}*2*sinx*cosx=0\ |:2\\\\cosx*(cosx+\sqrt{3} *sinx)=0\\\\cosx=0\\\\x_1=\frac{\pi }{2}+\pi n,\ \ \ n\in Z .\\\\cosx+\sqrt{3}*sinx=0\ |:2\\\\cosx*\frac{1}{2} +sinx*\frac{\sqrt{3} }{2} =0\\\\cosx*cos\frac{\pi }{3} +sinx*sin\frac{\pi }{3} =0\\\\cos(x-\frac{\pi }{3} )=0\\\\x-\frac{\pi }{3} =\frac{\pi }{2}+\pi n\\\\x_2=\frac{\pi }{2}+\frac{\pi }{3} +\pi n=\frac{5\pi }{6} +\pi n,\ \ \ n\in Z.[/tex]