Trig problem cos4x-7cos2x=8

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In summary, to solve the equation cos4x-7cos2x=8 in the interval [0,2\pi), we can use the double angle formula and the factoring method. However, in this case, there are no solutions as the resulting equation has no real solutions. The other factor, cos2x+1=0, has solutions at x=(pi)/2 and x=3(pi)/2. These are the only solutions in the given interval.
  • #1
Geekchick
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Homework Statement



cos4x-7cos2x=8 solve in interval [0,2[tex]\pi[/tex])

Homework Equations



reciprocal identities, pythagorean identities, sum/difference formulas, double angle formulas, power reducing formula, half-angle formula, product to sum formula, sum to product formula.

The Attempt at a Solution



So to solve this problem I am pretty sure I will have to use the double angle formula. However, I am a little confused as to how the formula changes when you have more than 2x. the only additional example my textbook gives it Cos6x=cos[tex]^{}2[/tex]3x-sin[tex]^{}2[/tex]3x. So from that I can assume that cos4x= cos[tex]^{}2[/tex]2x-sin[tex]^{}2[/tex]2x. But how would cos2x=2cos[tex]^{}2[/tex]x-1 change when its cos4x? I got my problem down to:

cos4x-14cos[tex]^{}2[/tex]x-15

I'm hoping that if I can expand on cos4x then I can maybe factor...or am I going about this all wrong.
Please help, I hate to be a bother but I'm taking this Pre-cal class independently over the summer and have already got an incomplete grade. My teacher said I have until the 20th to finish all the test and labs, so I'm trying to tie up all my loose ends so I can be done and move onto calculus.
Thanks
 
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  • #2
Geekchick said:
So to solve this problem I am pretty sure I will have to use the double angle formula. However, I am a little confused as to how the formula changes when you have more than 2x. the only additional example my textbook gives it Cos6x=cos[tex]^{}2[/tex]3x-sin[tex]^{}2[/tex]3x. So from that I can assume that cos4x= cos[tex]^{}2[/tex]2x-sin[tex]^{}2[/tex]2x. But how would cos2x=2cos[tex]^{}2[/tex]x-1 change when its cos4x? I got my problem down to:

cos4x-14cos[tex]^{}2[/tex]x-15

I'm hoping that if I can expand on cos4x then I can maybe factor...or am I going about this all wrong.

Ok well let's say we have [tex] cos(2y) = 2cos^2(y) - 1 [/tex]. Now to get [tex]cos(4x)[/tex], let [tex]y=2x[/tex]. It should make sense now. I used to have the same problem, which kind of goes back to functions. But if you think about it a little it shouldn't be hard to work out.
 
  • #3
I get what you are saying, but I still cannot solve the problem. :confused:
 
  • #4
You have the right idea. Write everything in terms of [tex]cos(2x)[/tex]and constants. Then see if you move everything one side and factor.
 
  • #5
Ok, so this is what I have done.

cos4x-7cos2x=8

seeing as cos4x=2cos22x-1

then

2cos22x-7co2x-9

When I factor that I get

(2cos2x-9)(cos2x+1)

Right so far?

then when I set each to zero I get

2cos2x=9

Here is where I'm a little confused. Does it go like this?

cos2x=9/2

cosx=9/4

or is it

cosx=9/2

all of these twos are throwing me lol


With the other one I get

cos2x=1
cosx=1/2

But another issue is when I plug the problem into my calculator and graph I end up with the answer being [tex]\pi[/tex]/2 and 3[tex]\pi[/tex]/2 implying that the answer is cosx=0 unless I am just having issues with my calculator which happens from time to time.

thanks
 
  • #6
Geekchick said:
When I factor that I get

(2cos2x-9)(cos2x+1)

Right so far?

OK up to here...

then when I set each to zero I get

2cos2x=9

Here is where I'm a little confused. Does it go like this?

cos2x = 9/2

This is correct. And this should instantly tell you that this case has no solution, as there is no angle (doubled or otherwise) which has a cosine larger than 1...

cosx=9/4

or is it

cosx=9/2

Neither of these is correct. It is not the case that cos(nx) = n cos(x) ; you cannot simply pull constant factors out of the argument of a trig function -- those constants are multiplying the angle and will not similarly multiply the trig function.

With the other one I get

cos2x=1
cosx=1/2

There is also a problem here. Your result for this case is (cos2x+1) = 0 , which gives

cos 2x = -1.

Again, you cannot just pull that '2' out of the argument of cosine: it is not something called cosine times 2x, but a function called cosine applied to the angle (2x).

The way you solve this is to ask what angles have a cosine of -1 (equivalently, we would solve 2x = arccosine(-1) ). We need all the angles in the interval [0, 2·pi), so we would look at

2x = (pi) , 2x = 3(pi), 2x = 5(pi), ... ,

which gives us

x = (pi)/2 , x = 3(pi)/2 [and that's it: everything beyond that is outside the specified domain]. Since the other factor gave no other solutions, these are the only ones.

As a check (which you should always do when working with trig equations), we find

cos4x - 7cos2x = 8

x = (pi)/2 : cos(2·pi) - 7·cos(pi) = 1 - 7·(-1) = 8 (check!)

x = 3(pi)/2 : cos(6·pi) - 7·cos(3·pi) = 1 - 7·(-1) = 8 (check!)
 

1. What is the equation cos4x-7cos2x=8 representing?

The equation represents a trigonometric problem that involves finding the solution for the value of x that satisfies the given equation.

2. How do you simplify cos4x-7cos2x=8?

To simplify this equation, we can use the double angle formula for cosine: cos2x = 2cos^2x - 1. Substituting this into the equation, we get cos4x - 7(2cos^2x - 1) = 8. From there, we can use algebraic methods to solve for x.

3. What are the possible solutions for cos4x-7cos2x=8?

The equation has an infinite number of solutions, as there are infinitely many angles that can satisfy the equation. However, the most commonly used solutions are typically in the range of 0 to 2π or 0 to 360 degrees.

4. How can I check if my solution for cos4x-7cos2x=8 is correct?

You can check your solution by substituting it back into the original equation and verifying that it satisfies the equation. You can also use a graphing calculator to graph the equation and see if your solution falls on the curve.

5. What are some real-world applications of solving trigonometric equations like cos4x-7cos2x=8?

Trigonometric equations are used in various fields such as astronomy, engineering, and physics to calculate distances, angles, and other measurements. In real-world scenarios, these equations can be used to determine the height of a building, the distance between two objects, or the angle at which a satellite needs to be launched for orbit.

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