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can someone help me solve cos(2x)= sqrt2/2?
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cos(2x) = \[\sqrt{2}/2\]
do you have the unit circle with you?
yeah
which angle gives you an x coordinate of sqrt(2)/2
pi/4 and 7pi/4
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so cos(pi/4) = sqrt(2)/2 and cos(7pi/4) = sqrt(2)/2
if you undo both sides with arccos, you get pi/4 = arccos(sqrt(2)/2) or 7pi/4 = arccos(sqrt(2)/2)
but its suppose to be cos (2x) = sqrt2/2
i know
so if you have cos(2x) = sqrt(2)/2 and you undo both sides with arccos, you get 2x = arccos(sqrt(2)/2))
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which means 2x = pi/4 or 2x = 7pi/4 solve each for x to get x = pi/8 or x = 7pi/8 this is assuming x is some angle between 0 and 2pi radians
thanks that makes sense
yw
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