Solve: \(\sqrt{2}secθ-tanθ=\sqrt{3}\)

1.	Solve: \(\sqrt{2}secθ-tanθ=\sqrt{3}\)
Answer» We have: \(\sqrt{2}secθ-tanθ=\sqrt{3}\) ⇒ \(\frac{\sqrt{2}}{cosθ}-\frac{sinθ}{cosθ}=\sqrt{3}\) ⇒ \(\sqrt{2}-sinθ =\sqrt{3}cosθ\) ⇒ \(\sqrt{3}cosθ+sinθ=\sqrt{2}\) Divide the equation by 2 \(\frac{\sqrt{3}}{2}cosθ+\frac{1}{2}sinθ=\frac{\sqrt{2}}{2}\) ⇒ \(cos(\frac{\pi}{6})cosθ+sin(\frac{\pi}{6})sinθ=\frac{1}{\sqrt{2}}\) ⇒ \(cos(θ-\frac{\pi}{6})=cos(\frac{\pi}{4})\) ⇒ \(θ-\frac{\pi}{6}=2n\pi ± \frac{\pi}{4}\) ⇒ \(θ=2n\pi ±\frac{\pi}{4}+\frac{\pi}{6}\) ⇒ \(θ=2n\pi +\frac{\pi}{4}+\frac{\pi}{6}\) or \(θ=2n\pi -\frac{\pi}{4}+\frac{\pi}{6}\) ⇒ \(θ=2n\pi +\frac{5\pi}{12}\) or \(θ=2n\pi-\frac{\pi}{12}\)

Answer»

We have: \(\sqrt{2}secθ-tanθ=\sqrt{3}\)

⇒ \(\frac{\sqrt{2}}{cosθ}-\frac{sinθ}{cosθ}=\sqrt{3}\)

⇒ \(\sqrt{2}-sinθ =\sqrt{3}cosθ\)

⇒ \(\sqrt{3}cosθ+sinθ=\sqrt{2}\)

Divide the equation by 2

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

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

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

⇒ \(θ-\frac{\pi}{6}=2n\pi ± \frac{\pi}{4}\)

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

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

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

Discussion