math
posted by gopi .
help me solve: find x in cos^2x  2 sinx cosx  sin^2x = 0

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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