Dividing both sides of the equation by
\(\sqrt{(\sqrt{3})^2 +(-1)^2}\) = \(\sqrt{4}\) = 2, we get \(\frac{\sqrt{3}}{2}\) sin θ - \(\frac{1}{2}\) cos θ = \(\frac{1}{\sqrt{2}}\)
⇒ cos 30° sin θ – sin 30° cos θ = \(\frac{1}{\sqrt{2}}\)
⇒ sin (θ – 30°) = sin \(\big(\) θ - \(\frac{π}{6}\) \(\big)\) = \(\frac{1}{\sqrt{2}}\)
⇒ sin \(\big(\) θ - \(\frac{π}{6}\) \(\big)\) = sin \(\frac{π}{4}\)
⇒ θ - \(\frac{π}{6}\) = nπ +(-1)n \(\frac{π}{4}\) , n ∈ I
⇒θ = nπ + \(\frac{π}{4}\) + (− 1)n \(\frac{π}{4}\) , n ∈ I .