number of solutions within 360 degrees

red and white kop!

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Jun 15, 2009
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Solve the following equation of A, giving solutions in the 0<A<360 inclusive (i don't know how to make the 'or equal to' sign):
sin2A - (sqrt3)(cos2A) =0
So I did it like this
Taking B=2A
sinB - (sqrt3)(sqrt(1-(sinB)^2)) = 0 (because cosB = sqrt(1-(sinB)^2)
sinB=sqrt(3-3(sinB)^2)
(sinB)^2=3-3(sinB)^2
4(sinB)^2=3
sinB=(plus or minus)sqrt3/2
from this I get the results A=30, 60, 120, 150, 210, 240, 300, 330
However, only 30, 120, 210 and 300 are correct.
Why is this?
 
red and white kop! said:
Solve the following equation of A, giving solutions in the 0<A<360 inclusive (i don't know how to make the 'or equal to' sign):
sin2A - (sqrt3)(cos2A) =0
So I did it like this
Taking B=2A
sinB - (sqrt3)(sqrt(1-(sinB)^2)) = 0 (because cosB = sqrt(1-(sinB)^2)
sinB=sqrt(3-3(sinB)^2)
(sinB)^2=3-3(sinB)^2
4(sinB)^2=3
sinB=(plus or minus)sqrt3/2
from this I get the results A=30, 60, 120, 150, 210, 240, 300, 330
However, only 30, 120, 210 and 300 are correct.
Why is this?

Whenever you square - and solve - you pull in chances of getting extraneous solutions.

another way to solve this:

sin(2A) - (?3)*(cos2A) =0

(1/2)*sin(2A) - (?3)/2*(cos2A) =0

cos(2A + ?/6) = cos[(2n+1)*?/2)]

2A = n? + ?/3 ? A = ?/6 , 2?/3, 7?/6 & 5?/3
 
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