Solve cos2x = 1 - sinx for 0° < x < 360° ?

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Answer 1 · 2018-04-27T06:16:15

The solutions are #S={kpi , 1/6pi+2kpi, 5/6pi+2kpi ; AA k in ZZ}#

The equation is

#cos2x=1-sinx#

#1-2sin^2x=1-sinx#

#2sin^2x-sinx=0#

#sinx(2sinx-1)=0#

#{(sinx=0),(2sinx-1=0):}#

#<=>#. #{(sinx=0),(sinx=1/2):}#

#<=>#, #{(x=kpi),(x=pi/6+2kpi), (x=5/6pi+2kpi):}#, #AA k in ZZ#