Trig equations

[paulxyz]

Hi, I am currently self-studying trig and working through text book questions. This question in the book, 2cos^3(Ø) = 3sinØcosØ was asked to solve for the interval -180<Ø<180. Eventually worked out that equation reduced to cosØ(2cos^2(Ø)-3sinØ) =0 and solved from there (answers +/- 90 deg and, 30 deg,150deg) However I first tried dividing through by cosØ in the original equation to give 2cos^2(Ø) = 3sinØ but his did not work and I was wondering why.

 

[Dr.Peterson]

However I first tried dividing through by cosØ in the original equation to give 2cos^2(Ø) = 3sinØ but this did not work and I was wondering why.

In doing that, you are essentially just ignoring the value of cosØ. Do you see that? It's as if you wrote cosØ(2cos^2(Ø)-3sinØ) =0 and just said, forget that first factor.

When you divide by something, you are assuming it is not zero. So here, you were assuming that cosØ is not zero, and so rejecting those solutions.

That's why you should always either use the factoring approach, or simply check separately whether what you divided by can be zero.

 

[khansaheb]

I was wondering why.

response #2 explained why you did not get the "complete" solution as suggested by your book.

Did you understand the response#2?

 

[jonhda1204200]

response #2 explained why you did not get the "complete" solution as suggested by your book.

Did you understand the response#2?

Can you answer this?

 

[khansaheb]

Can you answer this?

Yes I can. But first you need to share your work, so that we know why you are stuck

 

[prowesstom]

2cosθ(1-sin^2θ）=3sinθ.cosθ
cosθ（2-2sin^2θ-3sinθ）=0
cosθ=0 or 2-2sin^2θ-3sinθ
θ=+/-90 or θ=30/150

 

[The Highlander]

2cosθ(1-sin^2θ）=3sinθ.cosθ
cosθ（2-2sin^2θ-3sinθ）=0
cosθ=0 or 2-2sin^2θ-3sinθ
θ=+/-90 or θ=30/150

Welcome to the forum @prowesstom,

It can be a busy place at times, though, perhaps not as busy as it used to be.

Notwithstanding that, there is little point in dragging back up threads that involve queries that have already been addressed and answered satisfactorily.

Please read through a thread completely (especially old ones) before reviving it and adding information that is already in there (especially if that's just to demonstrate your own prowess in the subject matter).

If you have something new or informative to add to a thread then your contribution(s) would be most welcome but please consider whether it's really worth your time & effort to add anything to a thread like this (and our time & effort to read it) before posting.

Otherwise, I hope you enjoy your time here.

 

[matheww.parkerr]