-1/3 * (-√12*3) = -1/3 * (-√36) = -1/3 * (-6) = 2
Область допустимых значений уравнения : tgx≠0, х≠πn, n∈Z и х≠π/2+πk, k∈Z
![\frac{cos3x}{ \frac{sinx}{cosx} } =sin3x-2sinx, \\ cosx\cdot cos3x=sinx\cdot sin3x-2sin^{2}x, \\ cos3x\cdot cosx-sin3x\cdot sinx=-2sin^{2}x, \\ cos(3x+x)=-2sin^{2}x, \\ cos4x+2sin^{2}x=0, \\ (2cos ^{2} 2x-1)+(1-cos2x)=0, \\ 2cos ^{2} 2x-cos2x=0, \\ cos2x\cdot (2cos2x-1)=0 ](https://tex.z-dn.net/?f=+%5Cfrac%7Bcos3x%7D%7B+%5Cfrac%7Bsinx%7D%7Bcosx%7D+%7D+%3Dsin3x-2sinx%2C+%5C%5C++cosx%5Ccdot+cos3x%3Dsinx%5Ccdot+sin3x-2sin%5E%7B2%7Dx%2C+%5C%5C+cos3x%5Ccdot+cosx-sin3x%5Ccdot+sinx%3D-2sin%5E%7B2%7Dx%2C+%5C%5C+cos%283x%2Bx%29%3D-2sin%5E%7B2%7Dx%2C+%5C%5C++cos4x%2B2sin%5E%7B2%7Dx%3D0%2C+%5C%5C+%282cos+%5E%7B2%7D+2x-1%29%2B%281-cos2x%29%3D0%2C+%5C%5C+2cos+%5E%7B2%7D+2x-cos2x%3D0%2C+%5C%5C+cos2x%5Ccdot+%282cos2x-1%29%3D0%0A)
Произведение равно нулю, когда хотя один из множителей равен нулю, а другой при этом не теряет смысла.
![\left \{ {{cos2x=0} \atop {2cos2x-1=0}} \right. \Rightarrow \left \{ {{cos2x=0} \atop {cos2x= \frac{1}{2} }}\Rightarrow \left \{ {{2x= \frac{ \pi }{2} + \pi k,k\in Z} \atop {2x=\pm \frac{ \pi }{3}+2 \pi n,n\in Z }} \right. \right. \Rightarrow \left \{ {{x= \frac{ \pi }{4} + \frac{ \pi }{2}k,k\in Z } \atop {x=\pm \frac{ \pi }{6}= \pi n,n\in Z }} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bcos2x%3D0%7D+%5Catop+%7B2cos2x-1%3D0%7D%7D+%5Cright.+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Bcos2x%3D0%7D+%5Catop+%7Bcos2x%3D+%5Cfrac%7B1%7D%7B2%7D+%7D%7D%5CRightarrow+%5Cleft+%5C%7B+%7B%7B2x%3D+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B+%5Cpi+k%2Ck%5Cin+Z%7D+%5Catop+%7B2x%3D%5Cpm+%5Cfrac%7B+%5Cpi+%7D%7B3%7D%2B2+%5Cpi+n%2Cn%5Cin+Z+%7D%7D+%5Cright.++%5Cright.+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Bx%3D+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%2B+%5Cfrac%7B+%5Cpi+%7D%7B2%7Dk%2Ck%5Cin+Z+%7D+%5Catop+%7Bx%3D%5Cpm++%5Cfrac%7B+%5Cpi+%7D%7B6%7D%3D+%5Cpi+n%2Cn%5Cin+Z+%7D%7D+%5Cright.+)
Ответ.
![\frac{ \pi }{4}+ \frac{ \pi }{2} k, \pm \frac{ \pi }{6} + \pi n,k,n\in Z](https://tex.z-dn.net/?f=+%5Cfrac%7B+%5Cpi+%7D%7B4%7D%2B+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+k%2C++%5Cpm+%5Cfrac%7B+%5Cpi+%7D%7B6%7D+%2B+%5Cpi+n%2Ck%2Cn%5Cin+Z+)
Решение задания смотри на фотографии
1) 83²+33²+83*66 = 83²+2*83*33+33² = (83+33)² = 116²=13456
2) 19,3²+2*19,3*30,7+30,7² = (19,3+30,7)² = 50² = 2500
3) 31,8²-2*31,8*21,8+21,8² = (31,8-21,8)² = 10² = 100
4) 99*101² =(100-1)(100+1) = (100-1)(100²+200+1) = 99*10201=1009899
5) 201*199= (200+1)(200-1) = 200²-200+200-1 = 40000-1 = 39999
4 и 5 номера может у тебя неправильно переписаны?