Автор - mtolep

4Cos(x/2-pi/6)×dqrt12=0

Ответ

Автор - TimNeyd

4cos( frac{x}{2}+ frac{ pi }{6})+ sqrt{12}=0    


4cos( frac{x}{2}+ frac{ pi }{6})=- sqrt{12}   



cos( frac{x}{2}+ frac{ pi }{6})=- frac{ sqrt{12} }{4}  


cos( frac{x}{2} + frac{ pi }{6})=- frac{ sqrt{3} }{2}  


frac{x}{2} + frac{ pi }{6} =+- frac{5 pi }{6}+2 pi n,n


frac{x}{2}  =+- frac{5 pi }{6}- frac{ pi }{6} +2 pi n,n  


x=+- frac{10 pi }{6}- frac{ 2pi }{6} +4 pi n,n


x =+- frac{5 pi }{3}- frac{ pi }{3} +4 pi n,n

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