How do you solve #(2sinx-1)(2cos^2x-1)=0#?

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Answer 1 · 2018-06-25T06:41:56

The solutions are #S={pi/6+2kpi, 5/6pi+2kpi, pi/4+kpi/2}#, #AA k in ZZ#

The equation is

#(2sinx-1)(2cos2x-1)=0#

#=>#, #{(2sinx-1=0),(2cos^2x-1=0):}#

#=>#, #{(sinx=1/2),(cosx=+-sqrt(1/2)):}#

#=>#, # x=pi/6+2kpi " and " 5/6pi+2kpi #

and #x=pi/4+kpi/2#

#AA k in ZZ#

The solutions are

#S={pi/6+2kpi, 5/6pi+2kpi, pi/4+kpi/2}#