Solveeit Logo
Login

Question

Question: If \(3\theta = (2n + 1)\frac{\pi}{2};2\theta = (2n + 1)\frac{\pi}{2}\)where \(\theta = (2n + 1)\frac...

If 3θ=(2n+1)π2;2θ=(2n+1)π23\theta = (2n + 1)\frac{\pi}{2};2\theta = (2n + 1)\frac{\pi}{2}where θ=(2n+1)π2\theta = (2n + 1)\frac{\pi}{2}, then ⇒\Rightarrow

A

θ=30o,90o,150o,45o,135o\theta = 30^{o},90^{o},150^{o},45^{o},135^{o}

B

cosec θ+2=0\text{cosec }\theta + 2 = 0

C

⇒\Rightarrow

D

None of these

Answer

cosec θ+2=0\text{cosec }\theta + 2 = 0

Explanation

Solution

360∘360{^\circ}

⇒ ∴\therefore

⇒ θ=150∘\theta = 150{^\circ}

Neglecting (–) sign, we get

330∘330{^\circ} cos⁡2θ+sin⁡θ+1=0\cos^{2}\theta + \sin\theta + 1 = 0 1−sin⁡2θ+sin⁡θ+1=01 - \sin^{2}\theta + \sin\theta + 1 = 0.

The values of sin⁡2θ−sin⁡θ−2=0\sin^{2}\theta - \sin\theta - 2 = 0 between 0 and (sin⁡θ+1)(sin⁡θ−2)=0(\sin\theta + 1)(\sin\theta - 2) = 0 are sin⁡θ=2\sin\theta = 2.