Solve the following trig equation for all values on the interval of 0 le x < 2pi sin 4x=cos2x sinx(2)(2x)=cos2x 2sinx2x=cos2x 2sinx2x-cos2=0 cos2x(2sin-1) cos2x=0 and 2sinx=1

Raegan Bray

Raegan Bray

Answered question

2022-07-28

Solve the following trig equation for all values on the interval of 0 x 2 π
sin 4 x = cos 2 x
sin x ( 2 ) ( 2 x ) = cos 2 x
2 sin x 2 x = cos 2 x
2 sin x 2 x cos 2 = 0
cos 2 x ( 2 sin 1 )
cos 2 x = 0 and 2 sin x = 1

Answer & Explanation

Steven Bates

Steven Bates

Beginner2022-07-29Added 15 answers

sin 4 x = cos 2 x , 2 sin 2 x cos 2 x = cos 2 x
cos 2 x ( 2 sin 2 x 1 ) = 0
cos 2 x = 0 , sin 2 x = 1 / 2
1. cos 2 x = 0 , 2 x + π 2 + n π , x = π 4 + n π 2
x = π 4 , 3 π 4 , 5 π 4 , 7 π 4
2. 2 x = π 6 + 2 n π , 2 x = 5 π 6 + 2 n π
x = π 12 + n π , x = π 12 , 13 π 12
x = 5 π 12 + n π , x = 5 π 12 , 17 π 12
so, x = ( π 4 , 3 π 4 , 5 π 4 , 7 π 4 , π 12 , 5 π 12 , 13 π 12 , 17 π 12 )

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