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Question Number 17073 by Kunal kumar shukla last updated on 30/Jun/17

Commented by Kunal kumar shukla last updated on 30/Jun/17

$$wronganswer$$

Commented by Kunal kumar shukla last updated on 30/Jun/17

$$answer:-[2n\pi -\pi /6,2n\pi ]\cup [2n\pi +7\pi /6,2n\pi +2\pi ]$$

Commented by prakash jain last updated on 02/Jul/17

$${2\sin }^2\theta +3\sin \theta -2⩾0$$$$(2\sin \theta -1)(\sin \theta +2)⩾0$$$$\sin \theta +2\mathrm{is}\mathrm{always}+\mathrm{ve}$$$$2\sin \theta -1⩾0$$$$\Rightarrow \sin \theta ⩾\frac{1}{2}$$$$1stquadrant$$$$\theta \in [2n\pi +\frac{\pi }{6},2n\pi +\frac{\pi }{2}]$$$$2ndquadrant$$$$\theta \in [2n\pi +\frac{\pi }{2},2n\pi +\frac{5\pi }{6}]$$$$\theta \in [2n\pi +\frac{\pi }{6},2n\pi +\frac{5\pi }{6}]$$

Commented by 1234Hello last updated on 03/Jul/17

$$\mathrm{Does}\mathrm{the}\mathrm{answer}\mathrm{matches}?$$

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