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

2cos^2 x −25 cot x = 0 x =?

$$\:\:\:\:\:\mathrm{2cos}\:^{\mathrm{2}} {x}\:−\mathrm{25}\:\mathrm{cot}\:{x}\:=\:\mathrm{0}\: \\ $$$$\:\:\:\:\:{x}\:=? \\ $$

Answered by MJS_new last updated on 23/Oct/20

cos x =(1/( (√(1+tan^2 x)))) let t=tan x (2/(t^2 +1))−((25)/t)=0 t^2 −(2/(25))t+1=0 t=(1/(25))±((4(√(39)))/(25))i no real solution x=nπ+arctan ((1±4(√(39))i)/(25))

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

Commented by MJS_new last updated on 23/Oct/20

I forgot lim_(t→±∞) ((2/(t^2 +1))−((25)/t)) =0 ⇒ x=nπ±(π/2)

$$\mathrm{I}\:\mathrm{forgot}\:\underset{{t}\rightarrow\pm\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{25}}{{t}}\right)\:=\mathrm{0} \\ $$$$\Rightarrow\:{x}={n}\pi\pm\frac{\pi}{\mathrm{2}} \\ $$

Commented by benjo_mathlover last updated on 23/Oct/20

yes

$${yes} \\ $$

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

2cos^2 x −25 cot x = 0 ⇔2cos^2 x−25((cosx)/(sinx))=0 ⇒sinx≠0⇔x≠kπ ⇔cosx(2sinxcosx−25)=0 ⇔cosx(sin2x−25)=0(since sin2x−25)≠0 ⇔cosx=0⇔x=(π/2)+mπ(m∈Z)

$$\:\:\:\:\:\mathrm{2cos}\:^{\mathrm{2}} {x}\:−\mathrm{25}\:\mathrm{cot}\:{x}\:=\:\mathrm{0}\: \\ $$$$\Leftrightarrow\mathrm{2cos}^{\mathrm{2}} \mathrm{x}−\mathrm{25}\frac{\mathrm{cosx}}{\mathrm{sinx}}=\mathrm{0}\:\Rightarrow\mathrm{sinx}\neq\mathrm{0}\Leftrightarrow\mathrm{x}\neq\mathrm{k}\pi \\ $$$$\Leftrightarrow\mathrm{cosx}\left(\mathrm{2sinxcosx}−\mathrm{25}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{cosx}\left(\mathrm{sin2x}−\mathrm{25}\right)=\mathrm{0}\left(\mathrm{since}\:\mathrm{sin2x}−\mathrm{25}\right)\neq\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{cosx}=\mathrm{0}\Leftrightarrow\mathrm{x}=\frac{\pi}{\mathrm{2}}+\mathrm{m}\pi\left(\mathrm{m}\in\mathrm{Z}\right) \\ $$

Commented by MJS_new last updated on 23/Oct/20

you are right. but it′s even shorter cos x (2cos x −((25)/(sin x)))=0 ⇒ cos x =0

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}.\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{even}\:\mathrm{shorter} \\ $$$$\mathrm{cos}\:{x}\:\left(\mathrm{2cos}\:{x}\:−\frac{\mathrm{25}}{\mathrm{sin}\:{x}}\right)=\mathrm{0}\:\Rightarrow\:\mathrm{cos}\:{x}\:=\mathrm{0} \\ $$

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

But i want to explain that sin2x−25=0 has no real roots,Sir

$$\mathrm{But}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{explain}\:\mathrm{that}\:\mathrm{sin2x}−\mathrm{25}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{roots},\mathrm{Sir} \\ $$

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