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 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?

 

[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)

 

[steve snash]

ok thanks

 

[steve snash]

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

 

[icystrike]

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

 

[Mentallic]

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? I am really bad at trig identities =(

 

[rock.freak667]

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?

 

[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?

 

[Char. Limit]

Here's a clue...

Divide by 8 sin(x)...

 

[danago]

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!

 

[danago]

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.

 

[Char. Limit]

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

 

[danago]

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
<br /> 8 sin^2 (x) - sin(x) = 0<br />

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]

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