Trig Substitution

http://mathshelper.co.uk/STEP%20II%201989.pdf

I was doing the first question on that paper and I got upto solving cos(3x) = sqrt(3)/2 which has infinite solutions. My question is how do I know which 3 solutions would actually work in the original polynomial since the trig substitution gives infinite solutions.

(edited 13 years ago)

Reply 1

Farhan.Hanif93

18

Original post by tsrguru

http://mathshelper.co.uk/STEP%20II%201989.pdf

I was doing the first question on that paper and I got upto solving cos(3x) = sqrt(3)/2 which has infinite solutions. My question is how do I know which 3 solutions would actually work in the original polynomial since the trig substitution gives infinite solutions.

I haven't worked the question but I would have thought that you would take the principle values of theta (i.e. in the range

0\leq \theta \leq \dfrac{\pi}{2})

which give distinct values of

\cos \theta

as it's only the values of

\cos \theta

that you're actually looking for to be distinct, not

\theta

. Hope that makes sense?

EDIT: What I've written above is probably not too clear :p:

. Basically, there are an infinite number of solutions for theta, but only 3 distinct solutions for

\cos \theta

arise from those, you just need to identify which ones lead to distinct values of

\cos \theta

.

(edited 13 years ago)

Reply 2

DFranklin

18

Your solutions are going to be based on the values of cos x when cos(3x) = sqrt(3/2).

Although there are an infinite number of solutions to cos(3x)=sqrt(3/2), these will only give 3 different values for cos(x).

Reply 3

tsrguru

OP

Original post by DFranklin

Your solutions are going to be based on the values of cos x when cos(3x) = sqrt(3/2).

Although there are an infinite number of solutions to cos(3x)=sqrt(3/2), these will only give 3 different values for cos(x).

Thanks! that clarifies everything.

Trig Substitution

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