# number of solutions within 360 degrees

• May 17th 2011, 01:01 AM
furor celtica
number of solutions within 360 degrees
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?
• May 17th 2011, 05:38 AM
mr fantastic
Quote:

Originally Posted by furor celtica
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?

Extraneous solutions introduced by squaring and then square rooting. But why would you do it this way?

Much simpler to re-arrange:

$\sin(2A) = \sqrt{3} \cos(2A) \Rightarrow \tan(2A) = \sqrt{3}$ which is simple to solve.
• May 17th 2011, 05:53 AM
furor celtica
yes i thought it was that but wasn't sure how it happened.
good point with the simpler solution. not very clever of me.