# Homework Help: Finding intervals of trig functions

1. Mar 21, 2010

### steve snash

1. The problem statement, all variables and given/known data
solve the equation for x in the interval 0<=x<=2pi

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

3. 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?

2. Mar 21, 2010

### rock.freak667

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. Mar 21, 2010

### steve snash

ok thanks

4. Mar 22, 2010

### 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

5. Mar 22, 2010

### icystrike

$$cos(2x)=1-sin^{2}(x)$$ check your equation again

6. Mar 22, 2010

### Mentallic

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

7. Mar 22, 2010

### steve snash

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

8. Mar 22, 2010

### rock.freak667

right so bring the sinx on the left side and get

4sin2x-sinx=0

can you factor out sin(x) and get solutions?

9. Mar 22, 2010

### steve snash

simplifying trig functions

1. The problem statement, all variables and given/known data
4cos(2x)+sin(x)=4

in the interval 0<=x<=2pi

3. 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. Mar 22, 2010

### Char. Limit

Re: simplifying trig functions

Here's a clue...

Divide by 8 sin(x)...

11. Mar 22, 2010

### danago

Re: simplifying trig functions

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!

Last edited: Mar 22, 2010
12. Mar 22, 2010

### danago

Re: simplifying trig functions

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: Mar 22, 2010
13. Mar 22, 2010

### Char. Limit

Re: simplifying trig functions

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

14. Mar 23, 2010

### steve snash

Re: simplifying trig functions

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

15. Mar 23, 2010

### danago

Re: simplifying trig functions

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:

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

16. Mar 23, 2010