sin(cos3x-1/2)=0?

Question

what does x equal?
how would you solve this problem?

it doesnt make sense to me. help please!!!!

Answer 1

sin[cos(3x)-1/2]=0 --> let w =cos(3x)-1/2

then

sin[w]=0

taking the inverse of sin...arc sin...of both sides

arcsin(sin[w] = arcsin[0]

we get

[w] = arcsin[0] -----> [w] =0 //arcsin[0] =0

so

if [w] = 0 and

w=cos(3x)-1/2

then

cos(3x)-1/2 = 0 -----> solve this equation for which values

cos(3x)-1/2 = 0 then you got yourself the answer... you know what when does cos(3x) = 1/2? answer that

Answer 2

sin(cos3x-1/2)=0
so cos3x-1/2= K*pi K (any integer) But,as -1<=cos<=1
you can take only k=0
so for k=0
cos3x-1/2=0 cos3x=1/2 and 3x = pi/3+2npi and 3x=5pi/3+2npi
so x= pi/9+2npi/3 and x= 5pi/9+2npi/3

Answer 3

sin(cos 3x - 1/2) = 0
implies that cos 3x - 1/2 = 0
(or n pi, for n = 1, 2, 3, .. but cos z < 1)

we have cos 3x = 1/2, so 3x = 2n pi - pi/3 or 2n pi + pi/3, n = 0, 1, 2, ..
So x = pi/9, 5 pi/9, 7 pi/9, 11 pi/9, 13 pi/9, and 17 pi/9
(only considering values from 0 to 2pi)