Asked by gopi
help me solve: find x in cos^2x - 2 sinx cosx - sin^2x = 0
Answers
Answered by
Bosnian
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.