Asked by valeria hersh

4cos^2 theta = 3 where 0 degrees < theta< 360
Answered by Jai
4cos^2 theta = 3
cos^2 theta = 3/4
cos (theta) = sqrt(3) / 2
theta = arccos [ sqrt(3) / 2 ]

theta = π/6 , 11π/6

hope this helps~ XD
Answered by Jai
oops i forgot the +/- sign:

cos (theta) = +/- sqrt(3) / 2
theta = +/- arccos [ sqrt(3) / 2 ]

theta = π/6 , 11π/6 , 5π/6 , 7π/6

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