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

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

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

Answered by MJS last updated on 12/Jun/20

4cosec^2  x βˆ’9=cot x  9sin^2  x +sin x cos x βˆ’4=0  9cos 2x βˆ’sin 2x βˆ’1=0  x=arctan t  βˆ’((10t^2 +2tβˆ’8)/(t^2 +1))=0  t^2 +(1/5)tβˆ’(4/5)=0  t_1 =βˆ’1; t_2 =(4/5)  β‡’  x_(1.1) =((3Ο€)/4); x_(1.2) =((7Ο€)/4); x_(2.1) =arctan (4/5); x_(2.2) =Ο€+arctan (4/5)

$$\mathrm{4cosec}^{\mathrm{2}} \:{x}\:βˆ’\mathrm{9}=\mathrm{cot}\:{x} \\ $$ $$\mathrm{9sin}^{\mathrm{2}} \:{x}\:+\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:βˆ’\mathrm{4}=\mathrm{0} \\ $$ $$\mathrm{9cos}\:\mathrm{2}{x}\:βˆ’\mathrm{sin}\:\mathrm{2}{x}\:βˆ’\mathrm{1}=\mathrm{0} \\ $$ $${x}=\mathrm{arctan}\:{t} \\ $$ $$βˆ’\frac{\mathrm{10}{t}^{\mathrm{2}} +\mathrm{2}{t}βˆ’\mathrm{8}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$ $${t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{5}}{t}βˆ’\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{0} \\ $$ $${t}_{\mathrm{1}} =βˆ’\mathrm{1};\:{t}_{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{5}} \\ $$ $$\Rightarrow \\ $$ $${x}_{\mathrm{1}.\mathrm{1}} =\frac{\mathrm{3}\pi}{\mathrm{4}};\:{x}_{\mathrm{1}.\mathrm{2}} =\frac{\mathrm{7}\pi}{\mathrm{4}};\:{x}_{\mathrm{2}.\mathrm{1}} =\mathrm{arctan}\:\frac{\mathrm{4}}{\mathrm{5}};\:{x}_{\mathrm{2}.\mathrm{2}} =\pi+\mathrm{arctan}\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$

Answered by 1549442205 last updated on 12/Jun/20

we have (4/(sin^2 x))βˆ’9=((cosx)/(sinx))⇔4(1+cot^2 x)βˆ’9=cotx  4cot^2 xβˆ’cotxβˆ’5=0⇔(cotx+1)(4cotxβˆ’5)=0  a/cotx+1=0⇔cot x=βˆ’1=cot((3Ο€)/4)β‡’x∈{((3Ο€)/4);((7Ο€)/4)}  a/4cotxβˆ’5=0⇔cotx=(5/4)⇔tanx=(4/5)   β‡’x∈{arctan(4/5);arctan(4/5)+Ο€}  Thus,the roots in interval [0;2Ο€] of   given equation are:x∈{((3Ο€)/4);((7Ο€)/4);arctan(4/5);arctan(4/5)+Ο€}

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

Commented bybehi83417@gmail.com last updated on 12/Jun/20

thank you very much sir.

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

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