- 1 (31.06 KiB) Viewed 34 times
Solve: 4cos22x=1, where x is a real number x=(6n−4)(2π),n is an interger ;x=(6n−2)(2π),n is an interger x=(6n−4)(3π)
-
answerhappygod
- Site Admin
- Posts: 899604
- Joined: Mon Aug 02, 2021 8:13 am
Solve: 4cos22x=1, where x is a real number x=(6n−4)(2π),n is an interger ;x=(6n−2)(2π),n is an interger x=(6n−4)(3π)
Solve: 4cos22x=1, where x is a real number x=(6n−4)(2π),n is an interger ;x=(6n−2)(2π),n is an interger x=(6n−4)(3π),n is an interger ;x=(6n−2)(3π),n is an interger x=n(3π),n is an interger x=n(2π),n is an interger None of these
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!