what is the general solution of Cos(3)-Cos(5)=0?

Question

anonymous · Answer

is it
 $$\cos(3x)-\cos(5x)=0$$

anonymous · Answer

yes! the x is actually theta in the problem. I tried to use cosine of alpha + cosine of beta to solve.. but the answers i got were pretty strange looking

anonymous · Answer

$$\cos 3x=\cos 5x$$
$$5x=3x+2\pi k$$
$$2x=2\pi k$$
$$x=\pi k $$
and
$$5x=-3x+2\pi k$$
$$x=\frac{ \pi k}{4}$$

anonymous · Answer

Thank you! but did you divide the second solution by 5? how did you get the denominator to =4?

agent0smith · Answer

@Jonask good answer, but I didn't think it'd find all the solutions, since cosine is not a one-to-one function - you can't just drop the cosines from cos3x = cos5x (like you could with log 3x = log 5x since it is one to one). 3x does not necessarily have to be equal to 5x for their cosines to be equal (eg cos30 = cos330, which is not a multiple of 2k*pi.) http://www.wolframalpha.com/input/?i=%5Ccos+3x%3D%5Ccos+5x http://www3.wolframalpha.com/Calculate/MSP/MSP56361g05ba42ibdh1id700002aai2ee1dc288e24?MSPStoreType=image/png&s=28&w=317&h=1074

loser66 · Answer

what about this? $$cos (3x) - cos(5x) = 2sin\frac{3x+5x}{2}sin\frac{5x-3x}{2}$$=2sin(4x) sinx=0 iff x =0+ 2k$\pi$

anonymous · Answer

those answers make sense.. but since we're trying to be concise with the solutions. would i get  -pi+2piK? or -pi+piK? the period for cosine is 2pi.. im just wondering if the 2  cancels out.

anonymous · Answer

Thank you guys!