Find the general values of θ if (i) 2 sin θ – 1 = 0 (ii) cos θ

1.	Find the general values of θ if (i) 2 sin θ – 1 = 0 (ii) cos θ = \(-\frac{1}{2}\)(iii) 4 sin2θ = 1 (iv) tan 2x – \(\sqrt{3}\) = 0 (v) 2 cot2 θ = cosec2 θ
Answer» (i) 2 sin θ – 1 = 0 ⇒sin θ = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\) ⇒ θ = nπ + (–1)ⁿ\(\frac{\pi}{6}\) , n∈I (ii) cos θ = − \(\frac{1}{2}\) = cos (180° – 60°) = cos ( π - \(\frac{\pi}{3}\) ) = cos \(\frac{2\pi}{3}\) ∴ θ = 2nπ \(\pm\) \(\frac{2\pi}{3}\), n∈I (iii) 4 sin²θ = 1 ⇒ \(\frac{4(1-cos\,2\theta)}{2}\) = 1 ⇒ 2(1 - cos 2θ) =1 ⇒ 1-cos 2θ = \(\frac{1}{2}\) ⇒ cos 2θ = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\) ∴ 2θ = 2nπ ± \(\frac{\pi}{3}\) ⇒ θ = nπ ± \(\frac{\pi}{6}\) (iv) tan 2x – \(\sqrt{3}\) = 0 ⇒ tan 2x = \(\sqrt{3}\) = tan \(\frac{\pi}{3}\) ⇒ 2x = nπ + \(\frac{\pi}{3}\) ⇒ x = \(\frac{n\pi}{2}\) + \(\frac{\pi}{6}\) , n ∈ I (v) 2 cot²θ = cosec²θ ⇒ 2 cot²θ = 1 + cot²θ ⇒ cot²θ = 1 ⇒ cot θ = ± 1 = cot (± \(\frac{\pi}{4}\)) ⇒ θ = nπ ± \(\frac{\pi}{4}\) , n ∈ I

Find the general values of θ if (i) 2 sin θ – 1 = 0 (ii) cos θ = \(-\frac{1}{2}\)(iii) 4 sin2θ = 1 (iv) tan 2x – \(\sqrt{3}\) = 0 (v) 2 cot2 θ = cosec2 θ

Answer»

(i) 2 sin θ – 1 = 0

⇒sin θ = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)

⇒ θ = nπ + (–1)ⁿ\(\frac{\pi}{6}\) , n∈I

(ii) cos θ = − \(\frac{1}{2}\)

= cos (180° – 60°) = cos ( π - \(\frac{\pi}{3}\) ) = cos \(\frac{2\pi}{3}\)

∴ θ = 2nπ \(\pm\) \(\frac{2\pi}{3}\), n∈I

(iii) 4 sin²θ = 1

⇒ \(\frac{4(1-cos\,2\theta)}{2}\) = 1

⇒ 2(1 - cos 2θ) =1 ⇒ 1-cos 2θ = \(\frac{1}{2}\) ⇒ cos 2θ = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)

∴ 2θ = 2nπ ± \(\frac{\pi}{3}\)

⇒ θ = nπ ± \(\frac{\pi}{6}\)

(iv) tan 2x – \(\sqrt{3}\) = 0

⇒ tan 2x = \(\sqrt{3}\) = tan \(\frac{\pi}{3}\)

⇒ 2x = nπ + \(\frac{\pi}{3}\)

⇒ x = \(\frac{n\pi}{2}\) + \(\frac{\pi}{6}\) , n ∈ I

(v) 2 cot²θ = cosec²θ

⇒ 2 cot²θ = 1 + cot²θ

⇒ cot²θ = 1

⇒ cot θ = ± 1

= cot (± \(\frac{\pi}{4}\))

⇒ θ = nπ ± \(\frac{\pi}{4}\) , n ∈ I