solve the equation 5sin(θ -pi/6)=8cosθ for values of 0<=θ <=2pi

Answer 1

5sin(θ-π/6) = 8cosθ
5(sinθcosπ/6-cosθsinπ/6) = 8cosθ
5√3/2 sinθ - 5/2 cosθ = 8cosθ
5√3 sinθ = 21 cosθ
75 sin^2 θ = 441 cos^2θ
516 sin^2 θ = 441
sinθ = ±21/√516 = ±0.924
θ = 1.18 or 1.96 or 4.32 or 5.10
that is,
θ = .37π or .62π or 1.37π or 1.62π
so,
θ-π/6 = .21π or .46π or 1.21π or 1.46π

Now, we squared things to solve for θ, so we need to check for extraneous solutions. Note that the 2nd and 4th solutions are in QII and QIV, where sin and cos have opposite signs. So, only

θ = 1.18 or 4.32

To verify, see

http://www.wolframalpha.com/input/?i=solve+5sin%28%CE%B8-%CF%80%2F6%29+%3D+8cos%CE%B8+for+%CE%B8+%3D+0..2pi

Answer 2

or
from Steve's
5√3 sinθ = 21 cosθ

sinØ/cosØ = 21/5√3
tanØ = 21/5√3 , and the tangent is + in I and III

Ø = 1.18 in I
or
Ø = π + 1.18 = 4.32 in III