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#