How do you solve #(4cos^2theta - 2) + (6cos^2theta - 3) = 1#?

1 Answer
Alan P.
Aug 5, 2015

Within the range #theta epsilon [0,pi]#
#color(white)("XXXX")##theta = 0.684719# radians
or
#color(white)("XXXX")##theta = 2.456873# radians

Explanation:

#(4cos^2(theta) - 2)+ (6cos^2(theta)-3) = 1#

Is equivalent to
#color(white)("XXXX")##10cos^2(theta) = 6#

#rarr##color(white)("XXXX")##cos^2(theta) = 0.6#

#rarr##color(white)("XXXX")##cos(theta) =+- sqrt(0.6)#

#rarr##color(white)("XXXX")##theta = arccos(sqrt(0.6)) = 0.684719#
or
#rarr##color(white)("XXXX")##theta = arccos(-sqrt(0.6)) = 2.45873#

(Sorry, but I couldn't find anything prettier.)

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