I don't know how to do this exercise. Any help?

sin 3x-cosx/3=0
sin 3x=sin(2x+x)=sin2xcosx+cos2xsinx
sin2x=2sinxcosx
cos2x=1-2sin^2x
sin3x=3sinx-4sin^3x
That it's all I know. How can I transform cosx/3 to cos x?

CosmoLaura:

Please clarify your question, because \sin(3x) \ \ne \ \cos(x/3).

I've tried other combinations of notations trying to guess what you mean, thinking maybe you meant \sin^3x and (\cos x)/3, but none of it works out.

Sorry about that, but the exercise consist in solving this trigonometrical equation! Nothing else. I have to find a solution for x!

Ah, now I get it! I assuemd that this was an exercise to prove a trig identity - thanks for the clarification.

But I still am not sure - is the right hand side meant to be cos(x/3) or cos(x)/3?

cos(x/3)

Sorry of that. My fault. I didn't be attentive at this point when you wrote (cosx)/3.. It is cos(x/3).

OK, two ways to do this:

1. The brute force way (which quite honestlly was my first approach at this): use a numerical technique such as Newton's method to find 'x' that solves sin(3x)-cos(x/3) = 0.

or,

2. Try a trick. Given that cos(w) = sin(90-w), try substituting sin(90-x/3) for the right hand side. It works right out!

I get it! Thanks.