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)
Sqrt(x^2 + 4x + 4) - sqrt(x^2 - 6x + 9) = sqrt((x + 2)^2) - sqrt((x - 3)^2) = |x + 2| - |x - 3|
Если х = -7:
|-7 + 2| - |-7 - 3| = |-5| - |-10| = 5 - 10 = -5
Если х = -4:
|-4 + 2| - |-4 - 3| = |-2| - |-7| = 2 - 7 = -5
Ответ: -5
4√9-√65*√9+√65
12-2*(65)^1/2
36=6
2x - 4x = 0
-2x = 0
x = 0 : (-2)
x = 0
x≠0
8-x²/x=x при x≠0 8-x=x 2x=8 x=4
если неаккуратно написали и это (8-x²)/x=x
x≠0 8/x-x=x 2x=8/x 2x²=8 x²=4 x1=-2 x2=2