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2cos-2-x-25-cot-x-0-x-




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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