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}