View Full Version : Really hard Trignometric Equation!
 
 orkdork
Nov 23, 2009, 09:05 PM
csc^2 x/2=2 secx
 
it says to put all solutions in [0, 360) interval and to give all exact solutions (so they would end in 360k, where k is an interger).  I tried putting everything in sine and cosine, but it didn't work out :eek:
 KISS
Nov 23, 2009, 09:15 PM
Since you and I can't write in the math script that's possible on this board, at least use parenthesis and functions you might use in basic or Excel.
 
If that's csc(x/2)^2= 2*sec(x)
 
and it definitely looks like a mess:
 
Soln:  http://www.wolframalpha.com/input/?i=csc%28x%2F2%29^2%3D+2*sec%28x%29
 Unknown008
Nov 24, 2009, 07:47 AM
Hm, that's what I get:
 
cosec^2 (\frac{x}{2}) = 2 sec\ x
 
Convert all in cos and sin;
 
\frac{1}{sin^2 (\frac{x}{2})} = \frac{2}{cos\ x}
 
Cross multiply:
 
cos\ x = 2 sin^2 (\frac{x}{2})
 
Break cos into double angle using cos2A = 1 - 2sin^2A
 
We get:
 
1 -  2sin^2(\frac{x}{2})= 2 sin^2 (\frac{x}{2})
 
Putting all sin on one side, we get:
 
1= 4 sin^2 (\frac{x}{2})
 
\frac14= sin^2 (\frac{x}{2})
 
\sqrt{\frac14} = \frac12= sin (\frac{x}{2})
 
sin^{-1}(\frac12)= \frac{x}{2}
 
The angles for x/2 are 30 and 150 degrees. Then, for x become 60 and 300 degrees.
 
I hope it helped! :)
 galactus
Nov 24, 2009, 10:57 AM
Here is another way for kicks.
 
csc^{2}(\frac{x}{2})=\frac{2}{1-cos(x)}
 
Rewrite as 
 
\frac{2}{1-cos(x)}=\frac{2}{cos(x)}
 
2cos(x)=2-2cos(x)
 
cos(x)=\frac{1}{2}
 
x=\frac{\pi}{3}, \;\ \frac{5\pi}{3}