How do you solve the following equation #sin x cos x= 1/2# in the interval [0, 2pi]?

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Answer 1 · 2016-10-04T15:36:45

#S={pi/6, pi/3,(5pi)/6,(5pi)/3}#

#sinxcosx=1/2#

#sinx = 1/2, cosx=1/2#

#x=sin^-1(1/2)# OR #x=cos^-1(1/2)#

#x=pi/6 +2pin, x=(5pi)/6+2pin# OR #x=+-pi/3 +2pin#

#n=0, x=pi/6,(5pi)/6, pi/3,-pi/3#

#n=1, x=(13pi)/6, (17pi)/6,(7pi)/3,(5pi)/3#

#S={pi/6, pi/3,(5pi)/6,(5pi)/3}#