Ask Me Help Desk - Really hard Trignometric Equation!

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Really hard Trignometric Equation!

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:

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...D+2*sec%28x%29

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! :)

Here is another way for kicks.

$csc^{2}(\frac{x}{2})=\frac{2}{1-cos(x)}$

Rewrite as

$\frac{2}{1-cos(x)}=\frac{2}{cos(x)}$

$2cos(x)=2-2cos(x)$

$cos(x)=\frac{1}{2}$

$x=\frac{\pi}{3}, \;\ \frac{5\pi}{3}$

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