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Question: If \(\sin^{2}\theta + \sin\theta - 2 = 0 \Rightarrow (\sin\theta - 1)(\sin\theta + 2) = 0\), then th...

If sin⁡2θ+sin⁡θ−2=0⇒(sin⁡θ−1)(sin⁡θ+2)=0\sin^{2}\theta + \sin\theta - 2 = 0 \Rightarrow (\sin\theta - 1)(\sin\theta + 2) = 0, then the general value of ⇒\Rightarrowis.

A

sin⁡θ≠−2\sin\theta \neq - 2

B

∴sin⁡θ=1=sin⁡π/2\therefore\sin\theta = 1 = \sin\pi/2

C

⇒\Rightarrow

D

θ=nπ+(−1)nπ2\theta = n\pi + ( - 1)^{n}\frac{\pi}{2}

Answer

sin⁡θ≠−2\sin\theta \neq - 2

Explanation

Solution

tan⁡mθ=tan⁡nθ\tan m\theta = \tan n\theta

⇒mθ=pπ+nθ⇒θ=pπ(m−n)\Rightarrow m\theta = p\pi + n\theta \Rightarrow \theta = \frac{p\pi}{(m - n)}

⇒\Rightarrow πm−n\frac{\pi}{m - n}

tan⁡θ−2sec⁡θ=3⇒sin⁡θ−3cos⁡θ=2⇒\tan\theta - \sqrt{2}\sec\theta = \sqrt{3} \Rightarrow \sin\theta - \sqrt{3}\cos\theta = \sqrt{2} \Rightarrow or sin⁡(θ−π3)=sin⁡π4⇒θ=nπ+(−1)nπ4+π3\sin\left( \theta - \frac{\pi}{3} \right) = \sin\frac{\pi}{4} \Rightarrow \theta = n\pi + ( - 1)^{n}\frac{\pi}{4} + \frac{\pi}{3}.