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Mathematics 6 Online
OpenStudy (anonymous):

can someone help me solve cos(2x)= sqrt2/2?

OpenStudy (anonymous):

cos(2x) = \[\sqrt{2}/2\]

jimthompson5910 (jim_thompson5910):

do you have the unit circle with you?

OpenStudy (anonymous):

yeah

jimthompson5910 (jim_thompson5910):

which angle gives you an x coordinate of sqrt(2)/2

OpenStudy (anonymous):

pi/4 and 7pi/4

jimthompson5910 (jim_thompson5910):

so cos(pi/4) = sqrt(2)/2 and cos(7pi/4) = sqrt(2)/2

jimthompson5910 (jim_thompson5910):

if you undo both sides with arccos, you get pi/4 = arccos(sqrt(2)/2) or 7pi/4 = arccos(sqrt(2)/2)

OpenStudy (anonymous):

but its suppose to be cos (2x) = sqrt2/2

jimthompson5910 (jim_thompson5910):

i know

jimthompson5910 (jim_thompson5910):

so if you have cos(2x) = sqrt(2)/2 and you undo both sides with arccos, you get 2x = arccos(sqrt(2)/2))

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

thanks that makes sense

jimthompson5910 (jim_thompson5910):

yw

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