# Homework Help: Solving Trigonometric Equations

1. Dec 11, 2005

### opticaltempest

I have a question when solving trigonometric equations.

For example:

Find all the solutions in the interval [0,2pi)

$$\sin \theta \tan \theta = \sin \theta \]$$

If you choose to divide through by $$\sin \theta\]$$ we get,

$$\tan \theta = 1\]$$ such that $$\sin \theta \ne 0\]$$
otherwise we are essentially dividing both sides by zero, which we
cannot do.

Do we need to be careful when solving trigonometric equations using
multiplication and division?

We're dividing by a term that can take on the value of zero. Does
Are there any techniques to use when solving trig equations
so this doesn't happen?

2. Dec 11, 2005

### emptymaximum

you can devide through even if it has the posibility of being = to zero.
the reason for this is because trig functions are FUNCTIONS. when you take calculus you'll see that you can examin how a function behaves close to zero.
as φ -> 0 for sinφ/sinφ , that ratio actually -> 1.

when it is = 0 try not deviding it out.
also, recall tanφ = sinφ/cosφ, so tan(0) = 0/1 = 0.

3. Dec 11, 2005

### marlon

When you devide, you must make sure that the quantity that you use in the denominator is not zero. But in this case you can bypass division like this :

$$\sin \theta \tan \theta = \sin \theta$$
$$\sin \theta \tan \theta - \sin \theta = 0$$
$$\sin \theta ( \tan \theta -1) = 0$$

Then you solve both

$$\sin \theta = 0$$
and
$$\tan \theta -1 = 0$$

regards
marlon

4. Dec 11, 2005

### vaishakh

Squaring the terms is another thing which you have to be careful. You get extraneous roots out there.