Solve: \(\sqrt{3}cosθ+sinθ=\sqrt{2}\)

1.	Solve: \(\sqrt{3}cosθ+sinθ=\sqrt{2}\)
Answer» We have: \(\sqrt{3}cosθ+sinθ=\sqrt{2}\) Dividing both sides by 2, we get \(\frac{\sqrt{3}}{2}cosθ+\frac{1}{2}sinθ =\frac{\sqrt{2}}{2}\) ⇒ \(cosθ.cos\frac{\pi}{6}+sinv.sin\frac{\pi}{6}=\frac{1}{\sqrt{2}}\) ⇒ \(cos(θ-\frac{\pi}{6})=cos\frac{\pi}{4}\) ⇒ \(θ-\frac{\pi}{6}=2n\pi +\frac{\pi}{4},n∈z\) ⇒ \(θ=2n\pi +\frac{\pi}{4}+\frac{\pi}{6} n∈z\) ⇒ \(θ=2n\pi +\frac{\pi}{4}+\frac{\pi}{6}\) or \(θ=2n\pi -\frac{\pi}{4}+\frac{\pi}{6},n∈z\) ⇒ \(θ=2n\pi +\frac{5\pi}{12}\) or \(θ=2n\pi -\frac{\pi}{12},n∈z\)

Answer»

We have: \(\sqrt{3}cosθ+sinθ=\sqrt{2}\)

Dividing both sides by 2, we get

\(\frac{\sqrt{3}}{2}cosθ+\frac{1}{2}sinθ =\frac{\sqrt{2}}{2}\)

⇒ \(cosθ.cos\frac{\pi}{6}+sinv.sin\frac{\pi}{6}=\frac{1}{\sqrt{2}}\)

⇒ \(cos(θ-\frac{\pi}{6})=cos\frac{\pi}{4}\)

⇒ \(θ-\frac{\pi}{6}=2n\pi +\frac{\pi}{4},n∈z\)

⇒ \(θ=2n\pi +\frac{\pi}{4}+\frac{\pi}{6} n∈z\)

⇒ \(θ=2n\pi +\frac{\pi}{4}+\frac{\pi}{6}\) or \(θ=2n\pi -\frac{\pi}{4}+\frac{\pi}{6},n∈z\)

⇒ \(θ=2n\pi +\frac{5\pi}{12}\) or \(θ=2n\pi -\frac{\pi}{12},n∈z\)

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