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

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

Answered by liberty last updated on 29/Jun/21

$$\mathrm{since}\mathrm{the}\mathrm{equation}{\mathrm{x}}^2+4\mathrm{x}\cos \beta +\cot \beta =0$$$$\mathrm{has}\mathrm{multiple}\mathrm{roots},\mathrm{we}\mathrm{have}$$$$△=0\to 16\cos {}^2\beta -4\cot \beta =0$$$$\to 4\cot \beta (2\sin 2\beta -1)=0$$$$\mathrm{when}0<\beta <\frac{\pi }{2}\mathrm{we}\mathrm{get}$$$$\sin 2\beta =\frac{1}{2}\mathrm{so}\{\begin{array}{l}\beta =\frac{\pi }{12}\mathrm{or}\\ \beta =\frac{5\pi }{12}\end{array}.$$$$$$

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