Shoutbox

It's probably a very simple question but I can't see how to do it.

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Show that the equation

3tanX = 2cosX

can be expressed as

2sin²X + 3sinX - 2 = 0

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X = angle

   3tanX = 2cosX
->3(sinX/cosX) = 2cosX
->3sinX=2(cos²X)
->3sinX=2(1-sin²X)
->3sinX=2-2sin²X
->2sin²X+3sinX-2=0

(tanX=sinX/cosX and cos²X+sin²X=1)

Edit: to Mnjul - mine's formatted

And if two people do the same thing, it must be good

Thanks, someone PMed me it after I posted it too, it was identical to Mnjul's, he should have just posted in here

It makes sense now, god knows why I didn't think of moving over the cosX on the bottom

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Originally posted by raceprouk



Edit: to Mnjul - mine's formatted

And if two people do the same thing, it must be good

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what program creates those pictures?

the key to trig questions like that is memorising the list of trigonometric identities, seg.

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Originally posted by Time
the key to trig questions like that is memorising the list of trigonometric identities, seg.

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I knew them, but I have something wrong with me whereby I miss the simplest things.

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Originally posted by spokes

quote:

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Originally posted by raceprouk



Edit: to Mnjul - mine's formatted

And if two people do the same thing, it must be good

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what program creates those pictures?

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I use Equation Editor in Word, and take a screenshot