Posted by **gopi** on Saturday, January 12, 2013 at 11:14pm.

help me solve: find x in cos^2x - 2 sinx cosx - sin^2x = 0

- math -
**Bosnian**, Sunday, January 13, 2013 at 12:49am
cos ^ 2 ( x ) - sin ^ 2 ( x ) = cos ( 2x )

2 sin ( x ) cos ( x ) = sin ( 2 x )

Equation :

cos ^ 2 ( x ) - 2 sin( x ) cos( x ) -

sin ^ 2 ( x ) = 0

we can write like :

cos ( 2 x ) - sin ( 2 x ) = 0 Add sin ( 2 x ) to both sides

cos ( 2 x ) - sin ( 2 x ) + sin ( 2 x ) = 0 + sin ( 2 x )

cos ( 2 x ) = sin ( 2 x ) Divide both sides by cos ( 2 x )

cos ( 2 x ) / cos ( 2 x ) = sin ( 2 x ) / cos ( 2 x )

1 = tan ( 2 x )

tan ( 2 x ) = 1

tan ( pi / 4 ) = 1

Tangent is a periodic function with period pi , so :

2 x = n pi + pi / 4

( n is an integer ) Divide both sides by 2

x = n pi / 2 + pi / 8

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