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.
Trig equations
- Thread starter paulxyz
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Dr.Peterson
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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.