Question

How do you solve `cosx-4=sinx-4` for `0<=x<=2pi`?

Answer 1

`x= pi/4, x=(5pi)/4` for `0<= x <= 2pi`

Explanation:

`cosx-4=sinx-4 or cosx= sinx or tanx=1 ; tan(pi/4)=1 ; tan(pi+pi/4)=1 or tan((5pi)/4)=1 :. x= pi/4, x=(5pi)/4`[Ans]