How do you solve #\cos^2 x = \frac{1}{16} # over the interval #[0,2pi]#?

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Answer 1 · 2018-04-20T20:00:56

#x=cos^(-1)(1/4)=1.32 ["rad"]=75.5°#

#cos^2(x)=1/16#
#cos(x)=1/4#
#x=cos^(-1)(1/4)=1.32 ["rad"]=75.5#

Answer 2 · 2018-04-21T02:03:23

#x = +- 75^@52#

#cos^2 x = 1/16#
#cos x = +- 1/4#
Calculator and unit circle give -->
#x = +- 75^@52#