Question #63d49

1 Answer
Cem Sentin
Dec 5, 2017

#theta_1=pi/6#, #theta_2=(5pi)/6#

Explanation:

#cos(2theta)=3sin(theta)-1#

#1-2(sin(theta))^2=3sin(theta)-1#

#2(sin(theta))^2+3sin(theta)-2=0#

#(2sin(theta)-1)*(sin(theta)+2)=0#

From first multiplier, #sin(theta)=1/2#. Hence #theta_1=pi/6# and #theta_2=(5pi)/6#

No solution from second one.

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