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Trigonometric Functions @ganeshie8
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\[Solve~the~following~equation~for~0\le~x~\le360\]
\[\sin2x=1\]
The first value is 45 but i'm not sure how to find the second value
let u=2x then x=u/2 so you have 0<=u/2<=360 solving for u we get 0<=u<=720 so first step solve sin(u)=1 on the interval [0,720]
then once you find u then you can replace u with 2x and solve for x
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\[u=90\]\[x=\frac{ 90 }{ 2 }\]\[x=45\]is it correct? @freckles
ok but there are other u in the interval such that sin(u)=1
you have sin(u)=1 when u=90,90+360 and so 2x=90,90+360 so x=90/2 or x=(90+360)/2
@MARC_
\[x=45,225\]
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Thank you @freckles :)
np
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