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

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

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