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Question Number 22348 by Tinkutara last updated on 16/Oct/17

The sum of all the solutions of the equation 1 + 2 cosec x = −((sec^2 (x/2))/2) in the interval [0, 4π] is nπ, where n is equal to

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{1}\:+\:\mathrm{2}\:\mathrm{cosec}\:{x}\:=\:−\frac{\mathrm{sec}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}{\mathrm{2}}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{interval}\:\left[\mathrm{0},\:\mathrm{4}\pi\right]\:\mathrm{is}\:{n}\pi,\:\mathrm{where}\:{n}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to} \\ $$

Answered by ajfour last updated on 16/Oct/17

n=5 , as there are two possible solutions in the given interval. x=((3𝛑)/2) , ((7𝛑)/2) . Let tan (x/2) =t 2cosec x= (2/(sin x))=((1+t^2 )/t) sec^2 (x/2) =1+t^2 ; Then 1+((1+t^2 )/t)=−(((1+t^2 ))/2) 2t+2+2t^2 +t(1+t^2 ) =0 or t^3 +2t^2 +3t+2=0 ⇒ (t+1)(t^2 +t+2)=0 only one real root t=−1 ⇒ tan ((x/2))=−1 As 0 ≤ (x/2)≤ 2π ⇒ (x/2) = ((3π)/4) , ((7π)/4) ⇒ x=((3𝛑)/2) , ((7𝛑)/2) Their sum = 5𝛑 =n𝛑 So n=5 .

$$\:\:\:\:\:\:\boldsymbol{{n}}=\mathrm{5}\:,\:\:{as}\:{there}\:{are}\:{two} \\ $$$${possible}\:{solutions}\:{in}\:{the}\:{given} \\ $$$${interval}.\:\:\:\:\boldsymbol{{x}}=\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}}\:,\:\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{2}}\:. \\ $$$${Let}\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:={t} \\ $$$$\:\:\mathrm{2cosec}\:{x}=\:\frac{\mathrm{2}}{\mathrm{sin}\:{x}}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{{t}} \\ $$$$\:\:\:\:\:\mathrm{sec}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:=\mathrm{1}+{t}^{\mathrm{2}} \:\:;\:\:{Then} \\ $$$$\:\:\mathrm{1}+\frac{\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} }{\boldsymbol{{t}}}=−\frac{\left(\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} \right)}{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{2}{t}+\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} +{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:=\mathrm{0} \\ $$$${or}\:\:\:\boldsymbol{{t}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{t}}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{2}\right)=\mathrm{0} \\ $$$${only}\:{one}\:{real}\:{root}\:\boldsymbol{{t}}=−\mathrm{1} \\ $$$$\:\:\:\Rightarrow\:\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)=−\mathrm{1} \\ $$$$\:\:\:{As}\:\:\:\:\:\:\mathrm{0}\:\leqslant\:\frac{{x}}{\mathrm{2}}\leqslant\:\mathrm{2}\pi \\ $$$$\:\:\:\:\:\:\Rightarrow\:\frac{{x}}{\mathrm{2}}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:,\:\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:\:\:\boldsymbol{{x}}=\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}}\:,\:\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{2}}\: \\ $$$${Their}\:{sum}\:=\:\mathrm{5}\boldsymbol{\pi}\:=\boldsymbol{{n}\pi} \\ $$$$\:\:\:\:{So}\:\:\:\:\boldsymbol{{n}}=\mathrm{5}\:. \\ $$

Commented by Tinkutara last updated on 16/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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