Solve for the following equation in the interval [0, 2#pi#): #3cos^2theta + sin^2theta =2#?

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Answer 1 · 2017-12-16T01:07:35

#theta=pi/4, (3pi)/4, (5pi)/4, (7pi)/4#

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#3cos^2theta+sin^2theta=2#

#3cos^2theta+1-cos^2theta=2#

#2cos^2theta=1#

#cos^2theta=1/2#

#costheta=+-sqrt2/2#

#theta=pi/4, (3pi)/4, (5pi)/4, (7pi)/4#