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 Equations with Trig functions
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mathphobic
Average Member

USA
8 Posts

Posted - 10/23/2007 :  19:20:24  Show Profile  Reply with Quote
Can this be solved algebraically?

sqrt(3) * cos x + sin x = 1

I was asked to find the answers exactly, but I'm not sure how to do that with that equation.
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skeeter
Advanced Member

USA
5634 Posts

Posted - 10/23/2007 :  20:00:48  Show Profile  Reply with Quote
(3)*cosx + sinx = 1

(3)*cosx = 1 - sinx

square both sides ...

3cosx = 1 - 2sinx + sinx

use a well known identity ...

3(1 - sinx) = 1 - 2sinx + sinx

3 - 3sinx = 1 - 2sinx + sinx

0 = 4sinx - 2sinx - 2

0 = 2sinx - sinx - 1

0 = (2sinx + 1)(sinx - 1)

sinx = -1/2 ... x = 7/6, x = 11/6

sinx = 1 ... x = /2

use the original equation to check for possible extraneous solutions.
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mathphobic
Average Member

USA
8 Posts

Posted - 10/23/2007 :  21:29:14  Show Profile  Reply with Quote
Thank you skeeter. It never occurred to me to square both sides.
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 10/24/2007 :  07:44:15  Show Profile  Reply with Quote
quote:
Originally posted by mathphobic

Can this be solved algebraically?

sqrt(3) * cos x + sin x = 1

I was asked to find the answers exactly, but I'm not sure how to do that with that equation.



If you have an expression like

A * cos + B * sin - then

= (A + B)*{A/(A + B) * cos + B/(A + B) * sin}

= (A + B)* sin(+)

where

sin = A/(A + B) and

cos = B/(A + B)

for your problem

3 * cos(x) + 1 * sin(x) = 1

3/2 * cos(x) + 1/2 * sin(x) = 1/2

sin (x + /3) = sin (/6) or sin(-/6)

x = {/6 - /3} or {-/6 - /3}

and continue....
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