Solve the equation 2sin3x+sqrt3=0?

2 Answers
Mar 23, 2017

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

Explanation:

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,..............

Mar 24, 2017

see below

Explanation:

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}