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Find all solutions of trig

  • Thread starter jwxie
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
33,158
4,842

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
That's domain, and isn't relevant to this problem. You're not asked to find an inverse function, so the domain is all real numbers.
so 0<= 1/2du <= pi
and solve for du, which gives me 0 <= du <= 2pi
This is wrong. There's no calculus involved here; you're just asked to find the solutions of a trig equation.

You have cos(2x) = 0. This means that 2x = p/2, 3pi/2, 5pi/2, ..., -pi/2, -3pi/2, -5pi/2, ...
cos(u) = 0 when u is pi/2, 3pi/2, and etc.
Yes. And etc. includes the negative values I showed above.
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,
You aren't integrating, so there's no need for du.
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?

You're going at these about half wrong. All you need to do is solve the equations - there's no need for du, and there's no need to limit the domain. You are not integrating, and you are not finding an inverse that is a function.

If tan(5x) = 1, then 5x = pi/4, pi/4 + pi, pi/4 + 2pi, ... pi/4 - pi, pi/4 - 2pi, pi/4 - 3pi, ...

To solve for x, you're not dividing by 1/5 - you're dividing by 5.

Thanks
 
  • #3
280
0
Hi. thank you for your help.
It seems like I have to find the trend, which sometime is not very clear.
But I do understand why now.

Thanks.
 

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