1) 2SinxCosx = 6
Sin2x = 6
Решений нет, так как - 1 ≤ Sinx ≤ 1
![2)4SinxCosx=\sqrt{3}\\\\2Sin2x=\sqrt{3}\\\\Sin2x=\frac{\sqrt{3} }{2}\\\\2x=(-1)^{n}arcSin\frac{\sqrt{3} }{2} +\pi n,n\in Z\\\\2x=(-1)^{n}\frac{\pi }{3}+\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi }{6}+\frac{\pi n }{2},n\in Z](https://tex.z-dn.net/?f=2%294SinxCosx%3D%5Csqrt%7B3%7D%5C%5C%5C%5C2Sin2x%3D%5Csqrt%7B3%7D%5C%5C%5C%5CSin2x%3D%5Cfrac%7B%5Csqrt%7B3%7D+%7D%7B2%7D%5C%5C%5C%5C2x%3D%28-1%29%5E%7Bn%7DarcSin%5Cfrac%7B%5Csqrt%7B3%7D+%7D%7B2%7D+%2B%5Cpi+n%2Cn%5Cin+Z%5C%5C%5C%5C2x%3D%28-1%29%5E%7Bn%7D%5Cfrac%7B%5Cpi+%7D%7B3%7D%2B%5Cpi+n%2Cn%5Cin+Z%5C%5C%5C%5Cx%3D%28-1%29%5E%7Bn%7D%5Cfrac%7B%5Cpi+%7D%7B6%7D%2B%5Cfrac%7B%5Cpi+n+%7D%7B2%7D%2Cn%5Cin+Z)
3) 5tg²x - 4tgx - 1 = 0
Сделаем замену : tgx = m
5m² - 4m - 1 = 0
D = (-4)² - 4 * 5 * (- 1) = 16 + 20 = 36 = 6²
![m_{1}=\frac{4+6}{10}=1\\\\m_{2}=\frac{4-6}{10}=- 0,2\\\\tgx=1\\\\x=arctg1+\pi n,n\in Z\\\\x=\frac{\pi }{4} +\pi n,n\in Z\\\\tgx=-0,2\\\\x=arctg(-0,2)+\pi n,n\in Z\\\\x=-arctg0,2+\pi n,n\in Z](https://tex.z-dn.net/?f=m_%7B1%7D%3D%5Cfrac%7B4%2B6%7D%7B10%7D%3D1%5C%5C%5C%5Cm_%7B2%7D%3D%5Cfrac%7B4-6%7D%7B10%7D%3D-+0%2C2%5C%5C%5C%5Ctgx%3D1%5C%5C%5C%5Cx%3Darctg1%2B%5Cpi+n%2Cn%5Cin+Z%5C%5C%5C%5Cx%3D%5Cfrac%7B%5Cpi+%7D%7B4%7D+%2B%5Cpi+n%2Cn%5Cin+Z%5C%5C%5C%5Ctgx%3D-0%2C2%5C%5C%5C%5Cx%3Darctg%28-0%2C2%29%2B%5Cpi+n%2Cn%5Cin+Z%5C%5C%5C%5Cx%3D-arctg0%2C2%2B%5Cpi+n%2Cn%5Cin+Z)
4-5(3X+2.5)=3X+9.5
4-15х-12,5=3х+9,5
-15х-3х=9,5+12,5-4
-18х=18
х=18:(-18)
х=-1
Последний почему-то отличается
6х=2а-4
х=(2а-4)/6 по условию х>1
(2а-4)/6>1 2a-4>6 2a>10 a>5 При а>5 х>1
(2а-4)/4+6х=2а-5(х-1)
2а-4+24х=8а-20х+20
44х=6а+24
х=(6а+24)/44
(6а+24)/44< -1
6a+24 < -44
6a < -68
a <-10 1/3