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

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SUMMARY

The equation 20sin(x) - 4cos(x) - 13 = 0 is solvable by transforming it into a quadratic equation in terms of cos(x). By rewriting the equation as 20sin(x) = 4cos(x) + 13 and squaring both sides, one can derive a quadratic equation. Additionally, the equation can be expressed in the form sin(x + φ) = K, allowing for solutions when |K| ≤ 1. It is essential to verify the solutions against the original equation to determine their validity.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sin(x) and cos(x).
  • Familiarity with quadratic equations and their solutions.
  • Knowledge of the transformation of trigonometric equations into different forms.
  • Ability to verify solutions within the context of the original problem.
NEXT STEPS
  • Learn how to derive and solve quadratic equations from trigonometric identities.
  • Study the transformation of equations of the form A sin x + B cos x = C into sin(x + φ) = K.
  • Explore the implications of the identity sin^2(x) + cos^2(x) = 1 in solving trigonometric equations.
  • Review methods for verifying solutions in trigonometric contexts to ensure accuracy.
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Students studying trigonometry, educators teaching trigonometric equations, and anyone looking to enhance their problem-solving strategies in mathematics.

maphco
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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.
 
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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 =]
 
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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.
 

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