# Finding intervals of trig functions

• steve snash
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).

## 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?

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)

ok thanks

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

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

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

so 4sin^2(x)=sin(x) can be simplified to? I am really bad at trig identities =(

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?

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?

Here's a clue...

Divide by 8 sin(x)...

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

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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Char. Limit said:
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.

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

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

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

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

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

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