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ksaimouli
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\[0.2=\cos(\frac{ \pi t }{ 12 })\]
ksaimouli
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solve for t
Jonask
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\[\frac{\pi t}{12}=\arccos (0.2)\]
ksaimouli
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oopss i did not think this way thx
Jonask
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but are you allowed to use calculator \[\pi t/12=\pm \arccos(0.2)+2 \pi k,\forall k\in Z\]
ksaimouli
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i got 5.231 but i only got one what about other value
ksaimouli
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|dw:1359586504389:dw|
ksaimouli
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@Jonask
Jonask
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notice that there are infinite number of solutiond but the general solution of these
is
\[\huge t=\frac{12}{\pi}\arccos(0.2)+2\pi k\]
where k can be chosen from interger
k={....-2,-1,0,1,2,3,4...}
ksaimouli
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where that +2 pi k came from
Jonask
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this is for every period of cos wich is every 360 =2pi
but if you never did it just consider your answer + and -5.321
and 360-5.321
ksaimouli
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but the answers are +5.321 and 18.767
Jonask
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first \[\frac{t \pi}{12}=\cos^{-1}{0.2}=78\]