Problem in finding the General Solution of a Trigonometric Equation v2

[Wrichik Basu]

Homework Statement

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Find the general solution of the equation: $$\tan {x}+\tan {2x}+\tan {3x}=0$$

Answer given: ##x=## ##\frac {n\pi}{3}##, ##n\pi \pm \alpha## where ##\tan {\alpha} = \frac {1}{\sqrt {2}}##.

Homework Equations

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These equations may be used:

The Attempt at a Solution

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Please see the pic below:

The answer from the "EITHER" is correct, but how do I simplify the second part?

 

[BvU]

I find it hard to read and it is badly focused, so I am reluctant to type it out (after all you are too lazy to do so too ...), but I can distinguish you go from $$\left [ 1\ + \ {1\over 1-\tan\alpha\tan 2\alpha} \right ] = 0 $$ to something I don't understand and -- If I read it right -- don't believe...

Could you post the steps (typed) ?

 

[Wrichik Basu]

I find it hard to read and it is badly focused, so I am reluctant to type it out (after all you are too lazy to do so too ...), but I can distinguish you go from $$\left [ 1\ + \ {1\over 1-\tan\alpha\tan 2\alpha} \right ] = 0 $$ to something I don't understand and -- If I read it right -- don't believe...

Could you post the steps (typed) ?

I've posted a better picture. Please see. And the ##\alpha## in your post will be ##x##.

 

[BvU]

Better focused, yes. Understand or believe ?

Could you post the steps

 

[haruspex]

I've posted a better picture. Please see. And the ##\alpha## in your post will be ##x##.

Looks good as far as you went, but why did you stop?
You can throw away the denominator, and convert the cos 3x back to trig terms in x and 2x again.

 

[BvU]

Ah, the last ##\cos 3x## is more like ##\cos 3x## plus ...

 

[Wrichik Basu]

Looks good as far as you went, but why did you stop?
You can throw away the denominator, and convert the cos 3x back to trig terms in x and 2x again.

I could, but I would've also got a sin term, which I can take to other side, and then cross multiply to get tan terms in x and 2x. Then?

 

[haruspex]

I could, but I would've also got a sin term, which I can take to other side, and then cross multiply to get tan terms in x and 2x. Then?

You should get an equation involving tan x, tan 2x and a constant.

 

[Wrichik Basu]

You should get an equation involving tan x, tan 2x and a constant.

Yes, ##2\tan {x}\tan {2x}=0##.

Then...I understood. Thanks a lot.

 

[haruspex]

Yes, ##2\tan {x}\tan {2x}=0##.

Then...I understood. Thanks a lot.

did you mean, ##2-\tan {x}\tan {2x}=0##?

 

[Wrichik Basu]

did you mean, ##2-\tan {x}\tan {2x}=0##?

No, I did a wrong calculation. Then how will I proceed after that?

 

[BvU]

I second haru: ##2-\tan {x}\tan {2x}=0## to be solved . Repeat: see post #2.

 

[haruspex]

No, I did a wrong calculation. Then how will I proceed after that?

If you now have that equation, expand the tan 2x. If not, please post your working.