Как то так вроде:)
Удачи;)
Сейчас ещё попробую решить, а 3 в) я не знаю. А 4. вышло 0,5
![-1 \leq cos(\alpha) \leq 1](https://tex.z-dn.net/?f=-1+%5Cleq+cos%28%5Calpha%29+%5Cleq+1)
- возможные значения косинуса
![2cos^2(x) - 5cos(x) + 3 = 0\\\\ 2cos^2(x) - 2cos(x)-3cos(x) + 3 = 0\\\\ 2cos(x)*(cos(x) - 1)-3*(cos(x) -1) = 0\\\\ (2cos(x) - 3)*(cos(x) -1) = 0\\\\ 2cos(x)-3=0\ \ or\ \ cos(x)-1=0\\\\ cos(x)=\frac{3}{2}\ \ or\ \ cos(x)=1\\\\ cos(x)=1\\\\ x=2\pi n,\ n\in Z](https://tex.z-dn.net/?f=2cos%5E2%28x%29+-+5cos%28x%29+%2B+3+%3D+0%5C%5C%5C%5C%0A2cos%5E2%28x%29+-+2cos%28x%29-3cos%28x%29+%2B+3+%3D+0%5C%5C%5C%5C%0A2cos%28x%29%2A%28cos%28x%29+-+1%29-3%2A%28cos%28x%29+-1%29+%3D+0%5C%5C%5C%5C%0A%282cos%28x%29+-+3%29%2A%28cos%28x%29+-1%29+%3D+0%5C%5C%5C%5C%0A2cos%28x%29-3%3D0%5C+%5C+or%5C+%5C+cos%28x%29-1%3D0%5C%5C%5C%5C%0Acos%28x%29%3D%5Cfrac%7B3%7D%7B2%7D%5C+%5C+or%5C+%5C+cos%28x%29%3D1%5C%5C%5C%5C%0Acos%28x%29%3D1%5C%5C%5C%5C%0Ax%3D2%5Cpi+n%2C%5C+n%5Cin+Z)
Ответ:
![2\pi n,\ n\in Z](https://tex.z-dn.net/?f=2%5Cpi+n%2C%5C+n%5Cin+Z)
![\int { (\frac{6}{cos^2 \, 3x} +1 }) \, dx = 6*\int { \frac{1}{cos^2 \, 3x} } \, * \frac{1}{3}*d(3x) + \int { 1 } \, dx = \\ 2*tg(3x)+x+C](https://tex.z-dn.net/?f=+%5Cint+%7B+%28%5Cfrac%7B6%7D%7Bcos%5E2+%5C%2C+3x%7D+%2B1+%7D%29+%5C%2C+dx+%3D++6%2A%5Cint+%7B+%5Cfrac%7B1%7D%7Bcos%5E2+%5C%2C+3x%7D+%7D+%5C%2C+%2A+%5Cfrac%7B1%7D%7B3%7D%2Ad%283x%29+%2B+%5Cint+%7B+1+%7D+%5C%2C+dx+%3D+%5C%5C+2%2Atg%283x%29%2Bx%2BC)
Т.к. нам надо найти первообразную график которой проходит через точку M (π/4; 3π/4), то получаем, что:
3π/4=2*tg(3*π/4)+π/4+C
3π/4-π/4-2*(-1)=C
π/2+2=C
Получаем в итоге первообразную:
F(x)=2*tg(3x)+x+π/2+2