A.maths

sec^2 x = (2-cos x-sin x)/(1-sinx)
1 - sinx = ( 2 - cos x - sin x )( cos^2 x )
1 - sinx = ( 2 - cosx - sinx )( 1 - sin^2 x )
1 - sin x = ( 2 - cosx - sinx )( 1 + sin x )( 1 - sin x )
( 1 - sinx )[ 1 - ( 2 - cosx - sinx )( 1 + sin x ) ] = 0
( 1 - sin x )( 1 - 2 + cosx + cosxsinx - sinx + sin^2 x ) = 0
( 1 - sin x )( cosx + cosxsinx - sinx- cos^2x ) = 0
( 1 - sinx )[ cosx ( 1 - cosx ) - sinx ( 1 - cosx ) ] = 0
( 1 - sinx )( cosx - sinx )( 1 - cosx ) = 0
sinx = 1 or tanx = 1 or cosx = 1
x = 90*( rejected ), 45*, 225*, 0* or 360* for 0<= x <=360*
So x = 0*, 45*, 225* or 360*


2007-11-14 19:47:07 補充：
The range of values of x are not given, so I just presume it is 0 &lt;= x &lt;= 360* but in case it is not the case, say if it is 0 &lt;= x &lt;= 90*, then x = 0* or 45*. If it is 0 &lt; x &lt; 360*, then only 45* and 225* are the solutions.

2007-11-14 19:47:41 補充：
Last but not least, x = 90* has to be rejected since 1 / cos 90* = 1 / 0 which is undefined.

1/cos2x = 2-cosx-sinx/1-sinx
1/(1+sinx)(1-sinx)=2-cosx-sinx/1-sinx
1/1+sinx=2-cosx-sinx
2+2sinx-cosx-sinxcosx-sinx-sin2x=1
1+sinx-cosx-sinxcosx-sin2x=0
cos2x+sinx-cosx-sinxcosx=0
(cosx-1)(cosx-sinx)=0
cosx=1 or tanx=0
x=0 or 45