Thread: Trigonemetric Function - Solving

Hello,

Can someone please help me solve these 2 problems?

1. sqrt(3)sin2x=cos2x
2. cos(x/2)-1=0

Thanks!

Q1 Rewrite so you get sin2x/cos2x = 1/sqrt(3).
Let 2x = A and solve for A
Then put 2x back in and solve for x
Don't forget to consider the domain.

Q2 Let x/2 =A, write as cosA = 1, solve for A, sub in x/2 and solve for x. Again, watch the domain.

Originally Posted by Debsta

Q1 Rewrite so you get sin2x/cos2x = 1/sqrt(3).
Let 2x = A and solve for A
Then put 2x back in and solve for x
Don't forget to consider the domain.

Q2 Let x/2 =A, write as cosA = 1, solve for A, sub in x/2 and solve for x. Again, watch the domain.

Thanks! can you show me the algebraic step for Q2?

The answer is 4kpi...

mine looks like this:

cos(x/2)-1=0
cos(x/2)=1 which is pi +kpi
cos(x/2)= (pi + kpi)2
cosx=2pi+2kpi
is that right so far?

Originally Posted by l flipboi l

Thanks! can you show me the algebraic step for Q2?

The answer is 4kpi...

mine looks like this:

cos(x/2)-1=0
cos(x/2)=1 which is pi +kpi No, This means that x/2 = 2k pi, because
cos(even multiples of pi)=1 and cos(oddmultiples of pi) = -1. The 2 in the 2k pi means even multiples of pi
So x/2 = 2k pi
So x = 4k pi


cos(x/2)= (pi + kpi)2 Once you've brought in your pi's you don't have the cos there anymore.

cosx=2pi+2kpi and you can't say that if cos(x/2) = n then cos x = 2n You just can't take the 2 out like that.
is that right so far?

Hope that helps

Thanks!