Posted by jhay on .
1.) Find the solution set of
2sin squared θ1= 0 if 180 degrees is less than or equal to θ is less than or equal to 360 degrees
2.) prove:
2cotθ/cot squared θ 1 = sin2 θ sec2 θ

math 
Reiny,
2 sin^2 Ø = 1
sin^2 Ø = 1/2
sin Ø = 1/√2
if 180 ≤ Ø ≤ 360°
then Ø = 225° or 315°
2.
Did you mean:
2cotØ/(cot^2 Ø  1) = sin^2 Ø sec^2 Ø ?
If so, then it is not and identity, (I tried it with Ø=20°)
did you mean:
2cotØ/cot^2 Ø  1 = sin^2 Ø sec^2 Ø ?
If so, then it is not and identity, (I tried it with Ø=20°)
did you mean:
2cotØ/cot^2 Ø  1 = sin 2Ø sec 2Ø ?
If so, then it is not and identity, (I tried it with Ø=20°)
etc. 
math 
micheal,
4^(x 1)  6^x  2.9^(x 1) = 0