cos3x cosx + sin3x sinx = 0

[CrzyNic]

solving for 0<x>2pi

Cos3x * Cosx + Sin3x * Sinx = 0

I've no clue where to go on this one!!

 

[CrzyNic]

got some work on it...

cos3x= 4cos^3 x - 3cosx
sin3x= 3sinx - 4sin^3 x

so i'm here now:
(cosx * 4cos^3 x) - 4cosx + 4sinx - (4sin^3 x * sinx) = 0

 

[tkhunny]

What are you doing? Don't make it harder than it is. You should be familiar with as many identities as you can cram in your head.

Try this:

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

Ringing any bells?

 

[CrzyNic]

thanks

i wasn't seeing 3x and x as two diffrent angles....I found x to aproximatly pi/4

 

[tkhunny]

Pretty good, except why is it approximate and where are the other solutions?

Did you generate a numerical solution with a calculator? That is quite unnecessary.

cos(2x) = 0

Where is the cosine function zero (0)? pi/2, 3pi/2, 5pi/2, 7pi/2, etc...

cos(2x) = 0 leads to 2*x is in here ==> {pi/2, 3pi/2, 5pi/2, 7pi/2, etc...}
That puts x in here ==> {pi/4, 3pi/4, 5pi/4, 7pi/4, etc...}
Which ones are between 0 and 2pi?