Another trig question

This time i thought i got it right...

Solve the following

4tan \theta=tan2 \theta

.............

(0 \leq \theta <360)

My working..

4tan \theta=tan2 \theta, tan2 \theta = \frac{ 2tan \theta}{1- tan^2 \theta}

(\cdot ^\cdot \cdot)

...........

\frac{4tan \theta}{2tan \theta} =\frac{1}{1- tan^2 \theta}

(\cdot ^\cdot \cdot)

...............

2=\frac{1}{1- tan^2 \theta}

....

(\cdot ^\cdot \cdot)

... 2-2tan^2 \theta = 1 finaly

\pm \sqrt \frac{1}{2}=tan \theta

.

My answers.

35.3, 144.7, 215.3, 324.7

The books answer 0, 35.3, 144.7, 180, 215.3, 324.7

the books working the same as mine but then

2tan \theta(1- tan^2 \theta)=tan

then..

tan \theta(2 -2tan^2 \theta -1)=0

. How is this possible?

(edited 13 years ago)

Reply 1

DFranklin

You're dividing by tan theta. This is not a good thing to do when tan theta = 0.

Reply 2

nuodai

You should avoid dividing through by trig functions (or indeed any function, unless you know for sure that it is never zero). Instead, factorise.

If you saw the equation

x^2=2x

you'd take the

2x

over and factorise it to give

x(x-2)=0

; if you divided by

x

instead then you'd get

x=2

, which is just one of the two possible solutions. It's exactly the same situation here.

Reply 3

Core

I guess thats another lesson learnt, how did they get this?

tan \theta(2 -2tan^2 \theta -1)=0

Reply 4

Core

actualy i se ehow they did this now, thank you both for your help.

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