- #1

- 282

- 0

## Homework Statement

Find all solutions:

[1] cos(2x) = 0

[2] tan(5x) = 1

## Homework Equations

## The Attempt at a Solution

For [1], 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 [2], 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?**

Thanks