Soloving sqrt(5)Cos(2x + 0.464)

  • Thread starter thomas49th
  • Start date
In summary, you can solve equations like this by finding the x-coordinates of the points of interest, and drawing the curves that represent those x-coordinates.
  • #1
thomas49th
655
0

Homework Statement



Solve for the interval 0 < x < pi/2

[tex]\sqrt{5}Cos(2x+0.464)[/tex]

Homework Equations





The Attempt at a Solution



Am i right in saying as cos oscillate between -1 and 1:

The maximum is sqrt(5) and occurs when (2x-0.464) = 1

Now i got 0.464/2 and (2pi - 0.463)/2
and the back of the book says 0.32 and 2.36

How do you solve equations like above :S

Thanks

yes my calculator is in radians.
 
Physics news on Phys.org
  • #2
FIrst of all, what is it that you are trying to solve?
 
  • #3
oops i forgot.

solve for x in the range stated above

Thanks ;)
 
  • #4
thomas49th said:

Homework Statement



Solve for the interval 0 < x < pi/2

[tex]\sqrt{5}Cos(2x+0.464)[/tex]

Homework Equations





The Attempt at a Solution



Am i right in saying as cos oscillate between -1 and 1:

The maximum is sqrt(5) and occurs when (2x-0.464) = 1

Now i got 0.464/2 and (2pi - 0.463)/2
and the back of the book says 0.32 and 2.36

How do you solve equations like above :S

Thanks

yes my calculator is in radians.
In order to solve any equation, you first have to have an equation. What equation are you talking about. Exactly what is the question you want to answer?

Yes, the maximum of the expression you give is sqrt(5) but it does not occur when 2x+ 0.464= 1, it occurs when 2x+0.464= 0 or a multiplie of 2pi which is, I presume, where you got the values of 0.464/2 and (2pi- 0.463)/2 (it should be -0.464/2= -0.232 and (2pi 0.463)/2). But what is the question to which your book says the answer is 0.32 and 2.36?
 
  • #5
oops i forgot.

solve for x in the range stated above

Thanks ;)
 
  • #6
The rest of the question though is:

I) Express 2Cos2x-Sin2x in the form RCos(2x+a) giving R exactly and a to 3dp
II) Hence find the x-coordinates of the points of interestction of C1 and C2 in the interval 0<=x=<pi. Give your answer as radians and to 2dp

Note that c1 and c2 are curves with the equations y = cos2x - 2sin²x and y = sin2x respectively.
 
  • #7
it' ok i got it. cos-1(1/sqrt(5))

easy peasy really :)
 

What is the formula for solving sqrt(5)Cos(2x + 0.464)?

The formula for solving sqrt(5)Cos(2x + 0.464) is:
sqrt(5) * cos(2x + 0.464) = 0
This equation can be solved using trigonometric identities and algebraic manipulation.

What is the range of values for x in sqrt(5)Cos(2x + 0.464)?

The range of values for x in sqrt(5)Cos(2x + 0.464) is all real numbers, as cos(2x + 0.464) can take on any value between -1 and 1. This means there are infinitely many solutions to the equation.

Can sqrt(5)Cos(2x + 0.464) have complex solutions?

Yes, sqrt(5)Cos(2x + 0.464) can have complex solutions. This occurs when the value inside the cosine function results in an angle greater than 90 degrees or less than -90 degrees. In these cases, the solutions will be complex numbers.

How can I graph sqrt(5)Cos(2x + 0.464)?

To graph sqrt(5)Cos(2x + 0.464), you can use a graphing calculator or software. Simply input the equation and set the x-axis to the desired range. The resulting graph will be a sinusoidal curve with an amplitude of sqrt(5) and a period of pi/2.

What are some real-life applications of sqrt(5)Cos(2x + 0.464)?

sqrt(5)Cos(2x + 0.464) has many real-life applications, particularly in physics and engineering. It can be used to model wave motion, such as sound waves or electromagnetic waves. It is also commonly used in AC circuit analysis and signal processing. Additionally, it can be used in celestial mechanics to calculate the position and velocity of a planet in orbit.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
878
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
18
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
283
  • Calculus and Beyond Homework Help
Replies
1
Views
913
  • Calculus and Beyond Homework Help
Replies
6
Views
237
  • Precalculus Mathematics Homework Help
Replies
25
Views
547
  • Calculus and Beyond Homework Help
Replies
4
Views
940
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top