Solve \(\sqrt{3}\) sin θ – cos θ = \(\sqrt{2}\).

1.	Solve \(\sqrt{3}\) sin θ – cos θ = \(\sqrt{2}\).
Answer» 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 .

Answer»

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 .

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