Find all solutions:
 cos(2x) = 0
 tan(5x) = 1
The Attempt at a Solution
For , this is my attempt:
let u = 2x, du = 2 dx, so 1/2 du = x
let me define a range, 0<= x <= pi
so 0<= 1/2du <= pi
and solve for du, which gives me 0 <= du <= 2pi
cos(u) = 0 when u is pi/2, 3pi/2, and etc.
cos(du) = 0, when pi/2, 3pi/2 and etc all divide by 2x
this gives me pi/4, 3pi/4, and etc.
it comes out to be n*pi/4, where n = odd
for , i used the same approach
tan(5x) = 1
let u = 5x, 1/5du = dx,
i define range of tan this 0 <= x < pi/2 (not including pi/2)
so that replace x by 1/5 du and solve for du, i get new range 0<= du < 5pi/2
for tan(u) = 1, i have possible u to be pi/4, 5pi/4
and again, divide these by 1/5, i get tan(du) = 1 when pi/20, 5pi/20
so i thought the answer would be n* pi/20
now before i checked the answer, i asked myself, what about those limitation?
tan has certain limitations in range, so it can't be just n * pi/20
but i didn't figure it out
when i checked the answer, it was pi/20 + n*pi/5
so how did he get the last addition?