Question

How would I solve cos (sin x) = 1?

Answer 1

Focus on \(\cos u=1\) first. What could make \(\cos\) yield \(1\)? Well, how about \(u=2\pi k\) for integer \(k\) -- right?

Answer 2

Now the question is, when does \(\sin x=2\pi k\)?

Answer 3

Well, observe that \(\sin x\in[-1,1]\) -- and \(-1\le 2\pi k\le 1\) only for \(k=0\), correct?

Answer 4

2pik should belong to [-1,1]

Answer 5

... so it boils down to \(\sin x=0\) -- can you solve this?

Answer 6

I'm still a bit confused :/

Answer 7

@RPguy in solving \(\cos(\sin x)=0\), we're first interested in what \(\sin x\) has to be to make \(\cos(\sin x)=0\). Once we figure that out, we have to figure out what \(x\) has to be to yield those possible values of \(\sin x\).

Answer 8

Okay, that makes more sense

Answer 9

@RPguy so we observe that to make \(\cos(\sin x)=0\) we'd need \(\sin x\) to be some integer multiple of \(2\pi\), right? convince yourself of this by looking at the unit circle. Now, what integer multiples in particular are possible here?

Answer 10

Think of the "range" of \(\sin x\) -- it yields numbers between \(-1\) and \(1\), inclusive, correct? What integer multiples of \(2\pi\) lie in this range? Once you determine that, figure out what values of \(x\) could yield those values of \(\sin x\) :-)

Answer 11

Okay, so basically \[\cos (\sin 0)=1\]

Answer 12

@RPguy the only integer multiple in our desired range is \(0\), hence our problem reduces to solving \(\sin x = 0\). Here, however, we have more flexibility; we find \(x\) can be any integer multiple of \(\pi\). Hence \(x=\pi n\) for integer \(n\).

Answer 13

Why wouldn't it be \[n2\pi\]

Answer 14

it'd depend on the value of "n" I guess
otherwise sin(x) is 0 at 0 + \(\bf k\pi\)

Answer 15

hmm, oh, cosine has to be hmmm 1

Answer 16

yeah, \(n2\pi\) looks good

Answer 17

@jdoe0001 Okay thanks :)

Answer 18

No, \(2\pi k\) misses all the odd multiples and is an incomplete solution.