# Finding intervals of trig functions

steve snash

## 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?

Homework Helper
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)

steve snash
ok thanks

steve snash
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

icystrike
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

$$cos(2x)=1-sin^{2}(x)$$ check your equation again Homework Helper
And then to make things more obvious, assign y=sin(x). You'll need to swap back later though.

steve snash
so 4sin^2(x)=sin(x) can be simplified to? im really bad at trig identities =(

Homework Helper
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?

steve snash
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???

Gold Member

Here's a clue...

Divide by 8 sin(x)...

Gold Member

## 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 $$0 \leq x \leq 2\pi$$

You have simplified the equation which is good, but you now need to find x. Maybe write it as $$8 sin^2 (x) - sin(x) = 0$$. How would you solve it from there? HINT: Try factorizing!

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Gold Member

Here's a clue...

Divide by 8 sin(x)...

You need to be careful with this though, because what if $$sin(x)=0$$?

Purely by inspection of the equation, i can see that x = 0 is a possible solution. If you divide everything by $$8 sin(x)$$ however, the equation becomes $$0.125 = sin(x)$$, but now x=0 is no longer a solution.

Last edited:
Gold Member

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

steve snash

would the factorized version look like this???
(8sin(x)+1)(sin(x)-1)=0

Gold Member

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
$$8 sin^2 (x) - sin(x) = 0$$

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

$$8 sin^2 (x) - sin(x) = sin(x)(8 sin (x) - 1) = 0$$

Homework Helper
And then to make things more obvious, assign y=sin(x). You'll need to swap back later though. 