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How to solve for General Solutions ( sin2x=cosx )
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hint : sin 2x = 2sin x cos x.
How did you know that from? IDs?
yup.
wait which ID is that
double angle ID
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oh rightt
I got x=pi/2,3pi/2,5pi/6,pi/6
wow, you know how to get it...thats correct but its in the range of 0 to 2pi only.
._.
general solution....means all....in terms of n....right ?
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ah yes, then you must add/subtract integer multiples of the period (2pi) to each
alright so what would the formula be
for the first: \[x = \frac{\pi}{2} \pm 2k\pi\] k being any integer similar for the rest
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