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November 1st, 2009, 08:36 AM

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Quote:

Originally Posted by raza9990

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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 08:52 AM.

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