Solve the equation #2sin3x+sqrt3=0#?

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Answer 1 · 2017-03-23T17:12:37

General solution is #x=(npi)/3-(-1)^npi/9#

As #2sin3x+sqrt3=0#

#sin3x=-sqrt3/2=-sin(pi/3)#

Hence General Solution is #3x=npi+(-1)^n(-pi/3)#

or #3x=npi-(-1)^npi/3#

and #x=(npi)/3-(-1)^npi/9#

and a few solutions using #n=0,1,2,3,4,........# are

#x=-pi/9,(4pi)/9,(5pi)/9,(4pi)/3,..............#

Answer 2 · 2017-03-24T14:51:38

see below

#2 sin 3x + sqrt 3 =0#

#2 sin 3x=-sqrt3#

#sin 3x=-sqrt3/2#

#3x = sin^-1 (-sqrt3/2)#

#3x=-pi/3+2pin, or 3x=-(2pi)/3+2pin#

#x=(-pi/3+2pin)(1/3), or x=(-(2pi)/3+2pin)(1/3)#

#x=-pi/9+(2pi)/3 n, or x=-(2pi)/9+(2pi)/3 n#

#S={-pi/9+(2pi)/3 n,-(2pi)/9+(2pi)/3 n}#