Solving Trigonometric Equations

[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
this have any special name? How can I learn more about this?
Are there any techniques to use when solving trig equations
so this doesn't happen?

 

[emptymaximum]

you can divide 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.

 

[marlon]

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
this have any special name? How can I learn more about this?
Are there any techniques to use when solving trig equations
so this doesn't happen?

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

 

[vaishakh]

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