for-what-value-of-x-0-lt-x-lt-2-4cosec-2-x-9-cotx-

Question Number 98191 by behi83417@gmail.com last updated on 12/Jun/20

for what value of :{x∣ 0<x<2𝛑}: 4cosec^2 x−9=cotx

$$\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\::\left\{\boldsymbol{\mathrm{x}}\mid\:\mathrm{0}<\boldsymbol{\mathrm{x}}<\mathrm{2}\boldsymbol{\pi}\right\}:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}\boldsymbol{\mathrm{cosec}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}−\mathrm{9}=\boldsymbol{\mathrm{cotx}} \\ $$

Answered by MJS last updated on 12/Jun/20

4cosec^2 x −9=cot x 9sin^2 x +sin x cos x −4=0 9cos 2x −sin 2x −1=0 x=arctan t −((10t^2 +2t−8)/(t^2 +1))=0 t^2 +(1/5)t−(4/5)=0 t_1 =−1; t_2 =(4/5) ⇒ x_(1.1) =((3π)/4); x_(1.2) =((7π)/4); x_(2.1) =arctan (4/5); x_(2.2) =π+arctan (4/5)

$$\mathrm{4cosec}^{\mathrm{2}} \:{x}\:−\mathrm{9}=\mathrm{cot}\:{x} \\ $$$$\mathrm{9sin}^{\mathrm{2}} \:{x}\:+\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{9cos}\:\mathrm{2}{x}\:−\mathrm{sin}\:\mathrm{2}{x}\:−\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{arctan}\:{t} \\ $$$$−\frac{\mathrm{10}{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{8}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{5}}{t}−\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{0} \\ $$$${t}_{\mathrm{1}} =−\mathrm{1};\:{t}_{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}.\mathrm{1}} =\frac{\mathrm{3}\pi}{\mathrm{4}};\:{x}_{\mathrm{1}.\mathrm{2}} =\frac{\mathrm{7}\pi}{\mathrm{4}};\:{x}_{\mathrm{2}.\mathrm{1}} =\mathrm{arctan}\:\frac{\mathrm{4}}{\mathrm{5}};\:{x}_{\mathrm{2}.\mathrm{2}} =\pi+\mathrm{arctan}\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$

Answered by 1549442205 last updated on 12/Jun/20

we have (4/(sin^2 x))−9=((cosx)/(sinx))⇔4(1+cot^2 x)−9=cotx 4cot^2 x−cotx−5=0⇔(cotx+1)(4cotx−5)=0 a/cotx+1=0⇔cot x=−1=cot((3π)/4)⇒x∈{((3π)/4);((7π)/4)} a/4cotx−5=0⇔cotx=(5/4)⇔tanx=(4/5) ⇒x∈{arctan(4/5);arctan(4/5)+π} Thus,the roots in interval [0;2π] of given equation are:x∈{((3π)/4);((7π)/4);arctan(4/5);arctan(4/5)+π}

$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{4}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}−\mathrm{9}=\frac{\mathrm{cosx}}{\mathrm{sinx}}\Leftrightarrow\mathrm{4}\left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \mathrm{x}\right)−\mathrm{9}=\mathrm{cotx} \\ $$$$\mathrm{4cot}^{\mathrm{2}} \mathrm{x}−\mathrm{cotx}−\mathrm{5}=\mathrm{0}\Leftrightarrow\left(\mathrm{cotx}+\mathrm{1}\right)\left(\mathrm{4cotx}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\mathrm{a}/\mathrm{cotx}+\mathrm{1}=\mathrm{0}\Leftrightarrow\mathrm{cot}\:\mathrm{x}=−\mathrm{1}=\mathrm{cot}\frac{\mathrm{3}\pi}{\mathrm{4}}\Rightarrow\mathrm{x}\in\left\{\frac{\mathrm{3}\pi}{\mathrm{4}};\frac{\mathrm{7}\pi}{\mathrm{4}}\right\} \\ $$$$\mathrm{a}/\mathrm{4cotx}−\mathrm{5}=\mathrm{0}\Leftrightarrow\mathrm{cotx}=\frac{\mathrm{5}}{\mathrm{4}}\Leftrightarrow\mathrm{tanx}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\:\Rightarrow\mathrm{x}\in\left\{\mathrm{arctan}\frac{\mathrm{4}}{\mathrm{5}};\mathrm{arctan}\frac{\mathrm{4}}{\mathrm{5}}+\pi\right\} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{roots}\:\mathrm{in}\:\mathrm{interval}\:\left[\mathrm{0};\mathrm{2}\pi\right]\:\mathrm{of}\: \\ $$$$\mathrm{given}\:\mathrm{equation}\:\mathrm{are}:\mathrm{x}\in\left\{\frac{\mathrm{3}\pi}{\mathrm{4}};\frac{\mathrm{7}\pi}{\mathrm{4}};\mathrm{arctan}\frac{\mathrm{4}}{\mathrm{5}};\mathrm{arctan}\frac{\mathrm{4}}{\mathrm{5}}+\pi\right\} \\ $$

Commented by behi83417@gmail.com last updated on 12/Jun/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir}. \\ $$

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