How do you solve $2sin^2x=sinx$ ?

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Answer 1 · 2016-06-16T15:24:22

$x = {2kpi }uu {pi+2kpi }uu {pi/6+2kpi }uu {(5pi)/6+2kpi }$
for ${k=0,pm1,pm2,...}$

$2sin^2(x)=sin(x) -> sin (x)(2 sin (x)-1)=0$

so the conditions are

${(sin(x) = 0), (2sin(x)-1=0) :}$

The solutions are

$x = {pi+2kpi }uu {2kpi }$ for ${k=0,pm1,pm2,...}$

and

$x = {pi/6+2kpi }uu {(5pi)/6+2kpi }$ for ${k=0,pm1,pm2,...}$

Finally

$x = {2kpi }uu {pi+2kpi }uu {pi/6+2kpi }uu {(5pi)/6+2kpi }$

for ${k=0,pm1,pm2,...}$

How do you solve 2sin^2x=sinx2sin^2x=sinx?