Question Number 17073 by Kunal kumar shukla last updated on 30/Jun/17
Commented by Kunal kumar shukla last updated on 30/Jun/17
$${wrong}\:{answer} \\ $$
Commented by Kunal kumar shukla last updated on 30/Jun/17
![answer:−[2nπ−π/6,2nπ]∪[2nπ+7π/6,2nπ+2π]](Q17078.png)
$${answer}:−\left[\mathrm{2}{n}\pi−\pi/\mathrm{6},\mathrm{2}{n}\pi\right]\cup\left[\mathrm{2}{n}\pi+\mathrm{7}\pi/\mathrm{6},\mathrm{2}{n}\pi+\mathrm{2}\pi\right] \\ $$
Commented by prakash jain last updated on 02/Jul/17
![2sin^2 θ+3sin θ−2≥0 (2sin θ−1)(sin θ+2)≥0 sin θ+2 is always +ve 2sin θ−1≥0 ⇒sin θ≥(1/2) 1st quadrant θ∈[2nπ+(π/6),2nπ+(π/2)] 2nd quadrant θ∈[2nπ+(π/2),2nπ+((5π)/6)] θ∈[2nπ+(π/6),2nπ+((5π)/6)]](Q17194.png)
$$\mathrm{2sin}^{\mathrm{2}} \theta+\mathrm{3sin}\:\theta−\mathrm{2}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{2sin}\:\theta−\mathrm{1}\right)\left(\mathrm{sin}\:\theta+\mathrm{2}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{sin}\:\theta+\mathrm{2}\:\mathrm{is}\:\mathrm{always}\:+\mathrm{ve} \\ $$$$\mathrm{2sin}\:\theta−\mathrm{1}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}{st}\:{quadrant} \\ $$$$\theta\in\left[\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{6}},\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{2}{nd}\:{quadrant} \\ $$$$\theta\in\left[\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{2}},\mathrm{2}{n}\pi+\frac{\mathrm{5}\pi}{\mathrm{6}}\right] \\ $$$$\theta\in\left[\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{6}},\mathrm{2}{n}\pi+\frac{\mathrm{5}\pi}{\mathrm{6}}\right] \\ $$
Commented by 1234Hello last updated on 03/Jul/17
$$\mathrm{Does}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{matches}? \\ $$