Finding intervals of trig functions

In summary, the question is asking you to solve for x within the given interval, where sin(x) is a periodic function and 8 sin^2 (x) - sin(x) = sin(x)(8 sin (x) - 1).
  • #1
steve snash
50
0

Homework Statement


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

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

The Attempt at a Solution


I don't 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?
 
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  • #2
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)
 
  • #3
ok thanks
 
  • #4
so i simplified it to 4-2sin^2x+sin(x)=4 now what do i do, i don't 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
steve snash said:
so i simplified it to 4-2sin^2x+sin(x)=4 now what do i do, i don't 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
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? I am really bad at trig identities =(
 
  • #8
steve snash said:
so 4sin^2(x)=sin(x) can be simplified to? I am 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


Here's a clue...

Divide by 8 sin(x)...
 
  • #11


steve snash said:

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!
 
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  • #12


Char. Limit said:
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


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


steve snash said:
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 don't 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 said:
And then to make things more obvious, assign y=sin(x). You'll need to swap back later though.

:devil:
 

1. What is the definition of an interval in trigonometric functions?

An interval in trigonometric functions is a range of values for the independent variable (usually x) that results in a specific range of values for the dependent variable (usually y). This range is typically represented as [a, b], where a and b are the starting and ending values of the interval, respectively.

2. How do I find the intervals of a trigonometric function?

To find the intervals of a trigonometric function, you first need to identify the period and amplitude of the function. The period is the distance between two consecutive peaks or troughs of the function, while the amplitude is the distance between the midline and the peak or trough. Then, you can use the period and amplitude to determine the x-values that will result in a specific range of y-values, using the formula x = kπ ± a, where k is an integer and a is the amplitude. These x-values form the intervals of the function.

3. How do I know if a trigonometric function is increasing or decreasing on a specific interval?

To determine if a trigonometric function is increasing or decreasing on a specific interval, you need to look at the sign of the derivative of the function on that interval. If the derivative is positive, the function is increasing on that interval. If the derivative is negative, the function is decreasing on that interval. You can also look at the slope of the function on the interval - if it is positive, the function is increasing, and if it is negative, the function is decreasing.

4. What is a critical point in a trigonometric function?

A critical point in a trigonometric function is a point on the graph where the slope of the function is equal to zero. This means that the function is neither increasing nor decreasing at that point. Critical points can also occur at the endpoints of intervals, where the function may change direction or have a discontinuity.

5. How do I use intervals to solve trigonometric equations?

To solve trigonometric equations, you need to first find the intervals on which the equation is valid. This means finding the x-values that will result in the specified y-values. Then, you can use these intervals to determine the possible solutions to the equation. You may also need to consider any restrictions on the domain of the function, such as when the function is undefined or has a vertical asymptote.

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