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#1

$Old$ November 1st, 2009, 09:11 AM

raza9990 $raza9990 is offline$

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$Default$ Trignomectric Expressions and Equations

solve it for 0<x<360
4 cos^2x-3=cosec220

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#2

$Old$ November 1st, 2009, 09:32 AM

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I believe that I've got the question noted wrong

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I think you want $4cos^{2}(x)-3=cosec(220)$ (see e^(i*pi) below)

The one I noted down is
$4cos^{2}(x-3)=cosec(220)$

Which goes on to have complex solutions - if any exist - which I didn't realise - so I hope you didn't use any of my garbage from before

.

Last edited by rain; November 1st, 2009 at 11:50 AM. Reason: Misinterpreting question and writing garbage :(

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$Old$ November 1st, 2009, 09:36 AM

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Quote:

Originally Posted by raza9990 $View Post$

solve it for 0<x<360
4 cos^2x-3=cosec220

$sin(220) < 0$

$4cos^2x - (3+cosec220) = 0$

Use the difference of two squares and let $u=\sqrt{3+cosec(220)}$

$(2cos(x)-u)(2cos(x)+u) = 0$

$cos(x) = \frac{1}{2}\sqrt{3+cosec(220)}$

or

$cos(x) = -\frac{1}{2}\sqrt{3+cosec(220)}$

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Last edited by e^(i*pi); November 1st, 2009 at 09:52 AM.

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