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Question Number 119220 by benjo_mathlover last updated on 23/Oct/20

$$2\cos {}^{2}x-25\cot x=0$$$$x=?$$

Answered by MJS_new last updated on 23/Oct/20

$$\cos x=\frac{1}{\sqrt{1+{\tan }^{2}x}}$$$$\mathrm{let}t=\tan x$$$$\frac{2}{t^2+1}-\frac{25}{t}=0$$$$t^2-\frac{2}{25}t+1=0$$$$t=\frac{1}{25}\pm \frac{4\sqrt{39}}{25}\mathrm{i}$$$$\mathrm{no}\mathrm{real}\mathrm{solution}$$$$x=n\pi +\arctan \frac{1\pm 4\sqrt{39}\mathrm{i}}{25}$$

Commented by MJS_new last updated on 23/Oct/20

$$\mathrm{I}\mathrm{forgot}\underset{t\to \pm \mathrm{\infty }}{\lim }(\frac{2}{t^2+1}-\frac{25}{t})=0$$$$\Rightarrow x=n\pi \pm \frac{\pi }{2}$$

Commented by benjo_mathlover last updated on 23/Oct/20

$$yes$$

Answered by 1549442205PVT last updated on 23/Oct/20

$$2\cos {}^{2}x-25\cot x=0$$$$\iff {2\cos }^2\mathrm{x}-25\frac{\mathrm{cosx}}{\mathrm{sinx}}=0\Rightarrow \mathrm{sinx}\ne 0\iff \mathrm{x}\ne \mathrm{k}\pi$$$$\iff \mathrm{cosx}(2\mathrm{sinxcosx}-25)=0$$$$\iff \mathrm{cosx}(\sin 2\mathrm{x}-25)=0(\mathrm{since}\sin 2\mathrm{x}-25)\ne 0$$$$\iff \mathrm{cosx}=0\iff \mathrm{x}=\frac{\pi }{2}+\mathrm{m}\pi (\mathrm{m}\in \mathrm{Z})$$

Commented by MJS_new last updated on 23/Oct/20

$$\mathrm{you}\mathrm{are}\mathrm{right}.\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{even}\mathrm{shorter}$$$$\cos x(2\cos x-\frac{25}{\sin x})=0\Rightarrow \cos x=0$$

Commented by 1549442205PVT last updated on 23/Oct/20

$$\mathrm{But}\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{explain}\mathrm{that}\sin 2\mathrm{x}-25=0$$$$\mathrm{has}\mathrm{no}\mathrm{real}\mathrm{roots},\mathrm{Sir}$$

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