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Finding intervals of trig functions

  • #1

Homework Statement


solve the equation for x in the interval 0<=x<=2pi

4cos(2x)+sin(x)=4

The Attempt at a Solution


I dont understand what the question is asking me to do? Where do I start and how can this equation be made into an appropriate equation so i can answer the question?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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They want you to solve for x. But since sinx and cosx are periodic functions, you will an infinite number of solutions. So They gave you a specific interval.

Start by trying to express cos(2x) in terms of sin(x)
 
  • #4
so i simplified it to 4-2sin^2x+sin(x)=4 now what do i do, i dont really know what my aim is, as in what do i want my equation to look like before i start to find the values in the interval
 
  • #5
446
1
so i simplified it to 4-2sin^2x+sin(x)=4 now what do i do, i dont really know what my aim is, as in what do i want my equation to look like before i start to find the values in the interval
Be careful with your parenthesis.:tongue:
[tex]cos(2x)=1-sin^{2}(x)[/tex] check your equation again :rolleyes:
 
  • #6
Mentallic
Homework Helper
3,798
94
And then to make things more obvious, assign y=sin(x). You'll need to swap back later though.
 
  • #7
so 4sin^2(x)=sin(x) can be simplified to? im really bad at trig identities =(
 
  • #8
rock.freak667
Homework Helper
6,230
31
so 4sin^2(x)=sin(x) can be simplified to? im really bad at trig identities =(
right so bring the sinx on the left side and get

4sin2x-sinx=0

can you factor out sin(x) and get solutions?
 
  • #9
simplifying trig functions

Homework Statement


4cos(2x)+sin(x)=4

in the interval 0<=x<=2pi

The Attempt at a Solution


i got this far

sin(x)=8 sin^2(x)

in the interval 0<=x<=2pi
what is the question asking and how do i simplify this equation???
 
  • #10
Char. Limit
Gold Member
1,204
14


Here's a clue...

Divide by 8 sin(x)...
 
  • #11
danago
Gold Member
1,122
4


Homework Statement


4cos(2x)+sin(x)=4

in the interval 0<=x<=2pi

The Attempt at a Solution


i got this far

sin(x)=8 sin^2(x)

in the interval 0<=x<=2pi
what is the question asking and how do i simplify this equation???
Im guessing that the question is asking you to solve the equation for x, where x lies within the interval [tex]0 \leq x \leq 2\pi[/tex]

You have simplified the equation which is good, but you now need to find x. Maybe write it as [tex]8 sin^2 (x) - sin(x) = 0[/tex]. How would you solve it from there? HINT: Try factorizing!
 
Last edited:
  • #12
danago
Gold Member
1,122
4


Here's a clue...

Divide by 8 sin(x)...
You need to be careful with this though, because what if [tex]sin(x)=0[/tex]?

Purely by inspection of the equation, i can see that x = 0 is a possible solution. If you divide everything by [tex]8 sin(x)[/tex] however, the equation becomes [tex]0.125 = sin(x)[/tex], but now x=0 is no longer a solution.
 
Last edited:
  • #13
Char. Limit
Gold Member
1,204
14


Just remember the x=0 solution... it's not that hard, really.
 
  • #14


would the factorized version look like this???
(8sin(x)+1)(sin(x)-1)=0
 
  • #15
danago
Gold Member
1,122
4


would the factorized version look like this???
(8sin(x)+1)(sin(x)-1)=0
Im not really sure how you came to that result. If you try to expand it out again, you dont get back to
[tex]
8 sin^2 (x) - sin(x) = 0
[/tex]

The way to factorize it would be to note that sin(x) is a common factor of both terms:

[tex]8 sin^2 (x) - sin(x) = sin(x)(8 sin (x) - 1) = 0[/tex]
 
  • #16
Mentallic
Homework Helper
3,798
94
And then to make things more obvious, assign y=sin(x). You'll need to swap back later though.
:devil:
 

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