Question

Find the general solution of the following equation....
$cot^{2}Ф +\frac{3}{sinФ} +3=0$

Answer 1

Question :–

▪︎ Find the general solution of the following equation –

$\\ \implies { \bold{cot^{2} \phi +\dfrac{3}{sin \phi} +3=0}}$

ANSWER :–

$\\ \implies { \bold{cot^{2} \phi +\dfrac{3}{sin \phi} +3=0}}$

• We know that –

$\\ \implies { \red{ \boxed{ \bold{cot^{2} \phi = \frac{ {cos}^{2} \phi }{ {sin}^{2} \phi } }}}}$

• So that –

$\\ \implies { \bold{ \dfrac{cos^{2} \phi}{ {sin}^{2} \phi} +\dfrac{3}{sin \phi} +3=0}}$

$\\ \implies { \bold{ cos^{2} \phi \: + 3 {sin} \phi +3 {sin}^{2} \phi =0}}$

• We should write this as –

$\\ \implies { \bold{ cos^{2} \phi \: + {sin}^{2} \phi + 3 \: {sin} \phi +2 {sin}^{2} \phi =0}}$

• Let's use –

$\\ \implies { \red{ \boxed{ \bold{cos^{2} \phi \: + {sin}^{2} \phi=1 }}}}$

$\\ \implies { \bold{ 1 + 3 \: {sin} \phi +2 {sin}^{2} \phi =0}}$

$\\ \implies { \bold{ 2 {sin}^{2} \phi + 3 \: {sin} \phi + 1 =0}}$

$\\ \implies { \bold{ 2 {sin}^{2} \phi + 2 \: {sin} \phi + sin \phi+ 1 =0}}$

$\\ \implies { \bold{ 2 {sin} \phi ( \: {sin} \phi + 1)+ 1(sin \phi+ 1) =0}}$

$\\ \implies { \bold{( 2 \: {sin} \phi + 1) ( \: {sin} \phi + 1) =0}}$

$\\ \implies { \bold{ {sin} \phi = - \frac{1}{2} \: , \: {sin} \phi = - 1 }}$

▪︎ If $\: { \bold{sin \theta = {sin} \phi }} \:$ then general solution –

$\\ \implies { \red{ \boxed{ \bold{ \theta = n \pi + {( - 1)}^{n} }}}}$

$\\ \: \: . \: \: { \pink{\bold{ \underline{ when \: \: sin( \phi) = - \frac{1}{2} } : - }}}$

$\\ \implies { \bold{ {sin} \phi = - \frac{1}{2} }}$

$\\ \implies { \bold{ {sin} \phi = - sin( \frac{ \pi}{6} ) }}$

$\\ \implies { \bold{ {sin} \phi = sin( \pi + \frac{ \pi}{6} ) }}$

$\\ \implies { \bold{ {sin} \phi = sin( \frac{ 7\pi}{6} ) }}$

$\\ \implies {\bold{ \phi = n\pi + ( - 1) ^{n}( \frac{7\pi}{6}) \: \: \: [Genral \: \: solution]}}$

$\\ \: \: . \: \: { \pink{\bold{ \underline{ when \: \: sin( \phi) = - 1 } : - }}}$

$\\ \implies { \bold{ {sin} \phi = - 1 }}$

$\\ \implies { \bold{ {sin} \phi = - sin( \frac{\pi}{2} ) }}$

$\\ \implies { \bold{ {sin} \phi = sin( \pi + \frac{\pi}{2} ) }}$

$\\ \implies { \bold{ {sin} \phi = sin( \frac{3\pi}{2} ) }}$

$\\ \implies {\bold{ \phi = n\pi + ( - 1) ^{n}( \frac{3\pi}{2}) \: \: \: [Genral \: \: solution]}}$