How to solve this equation in $RR$ ? $2sin^2x-sqrt(3)sin2x=0$

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Answer 1 · 2017-04-18T16:35:30

$x =( k pi) uu (pi/3+k pi)$ for $k in ZZ$

Using $sin(2x)=2sin(x)cos(x)$ and substituting

$2sin(x)(sin(x)-sqrt3 cos(x))=0$

then we have

$sin(x)=0->x=k pi$ with $k in ZZ$

and

$sin(x)-sqrt3 cos(x)->tan(x)=sqrt3->pi/3+k pi$ with $k in ZZ$

so the solution is for

$x =( k pi) uu (pi/3+k pi)$ for $k in ZZ$

How to solve this equation in RRRR? 2sin^2x-sqrt(3)sin2x=02sin^2x-sqrt(3)sin2x=0