Ok, working on a trig equation, have part of it down to:

I need the left side factored (so I can use the zero rule to find values for x) but I can't seem to get it. Can anyone else?

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- November 14th 2006, 01:59 PMsinewave85need factors part of trig eqation
Ok, working on a trig equation, have part of it down to:

I need the left side factored (so I can use the zero rule to find values for x) but I can't seem to get it. Can anyone else? - November 14th 2006, 02:19 PMThePerfectHacker
Note, the error

In fact it is,

The presense of the absolute value function overcomplicates things.

What you can do, is express it as, (add and subtract 1)

Thus,

If,

Then,

You have a quadradic equation.

Just remember to check solutions because it is necessary though not suffienct. - November 14th 2006, 02:28 PMSoroban
Hello, sinewave85!

Have you**ever**tried to factor something with a square root in it?

Quote:

Remember "radical equations" from Algebra 2?

We have: .

Square both sides: .

. . and we have: .

Quadratic Formula: .

. . and we have: .

Don't forget to check for extraneous roots.

[And you wonder why you couldn't factor it . . . ] - November 14th 2006, 03:11 PMsinewave85
Thanks for the help -- both of you! I got it worked out, and I got the right answers, along with a few false ones. Just in case you are currious, the original problem was sin2x = cos3x, and the answers for x between 0 and 180 degrees are 18, 90, and 160 degrees. I am not sure I solved it the best possible way, but I did get the right answers -- with a little help ;). Thanks again for getting me through my brain fart.