November 2021 0 11 Report
Решите прошу срочно 1. sinx=-1,[-2Π;3Π/2]
2. cosx=1/2, [-Π;5Π/2]
3. tgx=√3,[-5Π/2;2Π];
4. ctgx=-3,[-Π;5Π/2]
5. 2sin^2x-7sinx+5=0, [-5Π/2;Π]
Please enter comments
Please enter your name.
Please enter the correct email address.
You must agree before submitting.

Copyright © 2024 SCHOLAR.TIPS - All rights reserved.