Kolten Conrad

Kolten Conrad

Answered

2022-07-02

Solving cos 2 θ sin θ = 1 , θ [ 0 , 360 ]

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Answer & Explanation

iskakanjulc

iskakanjulc

Expert

2022-07-03Added 18 answers

Notice, we have
cos 2 θ sin θ = 1
1 sin 2 θ sin θ = 1
sin 2 θ sin θ = 0
sin 2 θ + sin θ = 0
sin θ ( sin θ + 1 ) = 0
sin θ = 0 θ = n ( 180 )
Where, n is any integer.
But for the given interval [ 0 , 360 ], substituting n=0,1,2, we get
θ = 0 , 180 , 360
Now,
sin θ + 1 = 0
sin θ = 1 θ = 2 n ( 180 ) 90
But for given interval [ 0 , 360 ], substituting n=1 we get
θ = 270
Hence, we have
θ = { 0 , 180 , 270 , 360 }

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2d3vljtq

2d3vljtq

Expert

2022-07-04Added 5 answers

cos 2 ( θ ) sin ( θ ) = cos 2 ( θ ) + sin 2 ( θ ) , (using Pythagorean Identity)
sin 2 ( θ ) + sin ( θ ) = 0
let x = sin ( θ )
x 2 + x = 0
x = 1 ± 1 2 4 ( 1 ) ( 0 ) 2 , (Quadratic Formula)
x=0, x=−1
substitute sin ( θ ) back in and solve
0 = sin ( θ ) θ = 0 , 180 , 360
1 = sin ( θ ) θ = 270
θ = { 0 , 180 , 270 , 360 }

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