#sin(3θ−π/6)=−(√3)/2#
From https://www.math10.com/en/geometry/trigonometry-and-geometry-conversions/trigonometry.html
the identity is as follows
sin(A - B) = sin A cos B - cos A sin B
#[Sin 3θ][Cos(π/6)] - [Cos 3θ][Sin(π/6)] = −(sqrt3)/2#
using special triangles,
# Cos(π/6) = (√3)/2# , #Sin(π/6) = 1/2#
Thus,
#[(sqrt3)/2][Sin 3θ] - (1/2)[Cos 3θ] = −(sqrt3)/2#
multiply both sides of the equation by 2
#(sqrt3)[Sin 3θ] - [Cos 3θ] = -sqrt3#
then # cos 3θ = sqrt(1- sin^2 3θ)#
#(sqrt3)[Sin 3θ] - sqrt(1- sin^2 3θ) = -sqrt3#
#(√3)[Sin 3θ] +sqrt3 =sqrt(1- sin^2 3θ)#
squaring both sides and simplifying
#3sin^2 3θ +6[Sin 3θ] +3 = 1- Sin^2 3θ#
transposing all the terms from the right side to the left side and then combining like terms. Simplifying
#2sin^2 3θ + 3[Sin 3θ] +1 = 0#
then let #Sin 3θ = x#
# 2 x^2 + 3x +1 = 0#
then factor
#(2x +1) ( x+1) = 0#
then #x= -(1/2)# and #x = -1#
but #Sin 3θ = x#
solving for #θ#
#θ = -π/18# or #θ= -π/6#
