Author 
Topic 

mathphobic
Average Member
USA
8 Posts 
Posted  10/23/2007 : 19:20:24

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. 


skeeter
Advanced Member
USA
5634 Posts 
Posted  10/23/2007 : 20:00:48

(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. 


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
Advanced Member
USA
9116 Posts 
Posted  10/24/2007 : 07:44:15

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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