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Question Number 144800 by imjagoll last updated on 29/Jun/21

Let β be an acute angle such that the equation x^2 +4xcos β+cot β=0 involving variable x has multiple roots. Then the measure of β in radians is __

$$\mathrm{Let}\:\beta\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{angle}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{4xcos}\:\beta+\mathrm{cot}\:\beta=\mathrm{0} \\ $$$$\mathrm{involving}\:\mathrm{variable}\:\mathrm{x}\:\mathrm{has}\:\mathrm{multiple} \\ $$$$\mathrm{roots}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{measure}\:\mathrm{of}\:\beta\:\mathrm{in} \\ $$$$\mathrm{radians}\:\mathrm{is}\:\_\_ \\ $$

Answered by liberty last updated on 29/Jun/21

since the equation x^2 +4x cos β+cot β=0 has multiple roots , we have △=0→16cos^2 β−4cot β=0 →4cot β(2sin 2β−1)=0 when 0<β<(π/2) we get sin 2β=(1/2) so { ((β=(π/(12)) or)),((β=((5π)/(12)))) :}.

$$\:\mathrm{since}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{4x}\:\mathrm{cos}\:\beta+\mathrm{cot}\:\beta=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{multiple}\:\mathrm{roots}\:,\:\mathrm{we}\:\mathrm{have} \\ $$$$\bigtriangleup=\mathrm{0}\rightarrow\mathrm{16cos}\:^{\mathrm{2}} \beta−\mathrm{4cot}\:\beta=\mathrm{0} \\ $$$$\rightarrow\mathrm{4cot}\:\beta\left(\mathrm{2sin}\:\mathrm{2}\beta−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{when}\:\mathrm{0}<\beta<\frac{\pi}{\mathrm{2}}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\:\mathrm{sin}\:\mathrm{2}\beta=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{so}\:\begin{cases}{\beta=\frac{\pi}{\mathrm{12}}\:\mathrm{or}}\\{\beta=\frac{\mathrm{5}\pi}{\mathrm{12}}}\end{cases}. \\ $$$$ \\ $$

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