# Tough trig equation

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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Dick
Homework Helper
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 :)

Gib Z
Homework Helper
http://en.wikibooks.org/wiki/Trigonometry:Trigonometric_Identities_Reference [Broken]

The first one in "Sum to product" should help in these questions =]

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dynamicsolo
Homework Helper
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.