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mathphobic
Average Member

USA
8 Posts

 Posted - 10/23/2007 :  19:20:24 Can this be solved algebraically?sqrt(3) * cos x + sin x = 1I was asked to find the answers exactly, but I'm not sure how to do that with that equation.

skeeter

USA
5634 Posts

 Posted - 10/23/2007 :  20:00:48 (3)*cosx + sinx = 1(3)*cosx = 1 - sinxsquare both sides ...3cosx = 1 - 2sinx + sinxuse a well known identity ...3(1 - sinx) = 1 - 2sinx + sinx3 - 3sinx = 1 - 2sinx + sinx0 = 4sinx - 2sinx - 20 = 2sinx - sinx - 10 = (2sinx + 1)(sinx - 1)sinx = -1/2 ... x = 7/6, x = 11/6sinx = 1 ... x = /2use the original equation to check for possible extraneous solutions.

mathphobic
Average Member

USA
8 Posts

 Posted - 10/23/2007 :  21:29:14 Thank you skeeter. It never occurred to me to square both sides.

Subhotosh Khan

USA
9117 Posts

 Posted - 10/24/2007 :  07:44:15 quote:Originally posted by mathphobicCan this be solved algebraically?sqrt(3) * cos x + sin x = 1I was asked to find the answers exactly, but I'm not sure how to do that with that equation.If you have an expression likeA * cos + B * sin - then= (A + B)*{A/(A + B) * cos + B/(A + B) * sin}= (A + B)* sin(+)wheresin = A/(A + B) andcos = B/(A + B)for your problem3 * cos(x) + 1 * sin(x) = 13/2 * cos(x) + 1/2 * sin(x) = 1/2sin (x + /3) = sin (/6) or sin(-/6)x = {/6 - /3} or {-/6 - /3}and continue....
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