Solve Trig Equation 20sinx - 4cosx - 13 = 0

  • Thread starter Thread starter maphco
  • Start date Start date
  • Tags Tags
    Trig
AI Thread Summary
The equation 20sinx - 4cosx - 13 = 0 is solvable by transforming it into a quadratic equation for cos(x). By rewriting the equation as 20sinx = 4cosx + 13 and applying the identity sin(x) = sqrt(1 - cos(x)^2), one can square both sides to derive the quadratic form. Additionally, the equation can be expressed in the form A sin x + B cos x = C, which allows for further manipulation using trigonometric identities. It is crucial to verify the solutions against the original equation to determine their validity. This approach provides a systematic strategy for solving similar trigonometric equations.
maphco
Messages
24
Reaction score
0
20sinx - 4cosx - 13 = 0 : solvable?

Is this equation solvable? I've been banging my head on it for the past half hour and don't know how to start.

20sinx - 4cosx - 13 = 0

Also, this started out as a physics question, but seeing as this is just math stuff now I thought I'd put it here. I know all the physics was done correctly, so this is how it should end up.
 
Last edited:
Physics news on Phys.org
Yes, you can solve it. Write sin(x)=sqrt(1-cos(x)^2). Now write the equation as 20sinx=4cosx+13 and square both sides. You should get a quadratic equation for cos(x). Solve it.
 
Oh wow, that's brilliant! Thanks, Dick. I'm going to put that in my trig solving strategy log :)
 
http://en.wikibooks.org/wiki/Trigonometry:Trigonometric_Identities_Reference

The first one in "Sum to product" should help in these questions =]
 
Last edited by a moderator:
It should be mentioned that the two methods, which are essentially different ways of making the same manipulation, will give you two answers: the first because it yields a quadratic equation, the second because equations of the form

A sin x + B cos x = C

can be transformed (using the identity GibZ refers to) into the form

sin (x + phi) = K

and can be solved if |K|<=1 .

You will need to check the results against your original problem to see which solution (or perhaps both) applies to the situation.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top