Solve the following trig equation for all values on the interval of: 0 le x < 2pi sin4x=cos2x sin(2)(2x)=cos2x 2sinx2x=cos2x -cos2x -cos2x =2sin2x-cos2x=0 cos2x(2sin-1) cos2x=0 and 2sinx-1=0

beatricalwu 2022-08-01 Answered
Solve the following trig equation for all values on the interval of: 0 x 2 π
sin 4 x = cos 2 x
sin ( 2 ) ( 2 x ) = cos 2 x
2 sin x 2 x = cos 2 x
cos 2 x cos 2 x
= 2 sin 2 x cos 2 x = 0
cos 2 x ( 2 sin 1 )
cos 2 x = 0 and 2 sin x 1 = 0
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Answers (1)

yelashwag8
Answered 2022-08-02 Author has 17 answers
sin 4 x = cos 2 x
sin ( 2.2 x ) = cos 2 x
2 sin ( 2 x ) . cos ( 2 x ) = cos ( 2 x )
2 sin ( 2 x ) . cos ( 2 x ) cos ( 2 x ) = 0
cos ( 2 x ) . ( 2 sin ( 2 x ) 1 ) = 0
cos ( 2 x ) = 0   o r   ( 2 sin ( 2 x ) 1 ) = 0
cos ( 2 x ) = cos ( π / 2 )   o r   2 sin ( 2 x ) = 1
2 x = π / 2   o r   sin ( 2 x ) = 1 / 2
x = π / 4   o r   s i n ( 2 x ) = sin ( π / 6 ) x = π / 12
cos ( 2 x ) = 0   o r   sin ( 2 x ) = sin ( 5 π / 6 ) x = 5 π / 12
cos ( 2 x ) = cos ( 3 π / 2 )
2 x = 3 π / 2
x = 3 π / 4
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