Math, asked by tanishkajadhav982, 3 months ago

Find the principal solution of √3 cosec x + 2 = 0​

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Answers

Answered by siddhikatathe70
1

Step-by-step explanation:

option c is correct answer thank you

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Answered by mathdude500
3

\large\underline{\sf{Solution-}}

Concept Used :-

Principal solution of Trigonometric Equation :-

  • The solution of Trigonometric Equation lies in the interval [0, 2 π) is called Principal Solution.

Given that,

\rm :\longmapsto\: \sqrt{3} cosecx + 2 = 0

\rm :\longmapsto\: \sqrt{3} cosecx  =  - 2

\rm :\longmapsto\:cosecx \:  =  -  \: \dfrac{2}{ \sqrt{3} }

\rm :\longmapsto\:sinx \:  =  -  \: \dfrac{ \sqrt{3} }{2}

We know,

 \purple{\rm :\longmapsto\:sinx < 0 \: when \: x \: lies \: in \:  {3}^{rd}  \: and \:  {4}^{th} \: quadrant.}

and

 \purple{\rm :\longmapsto\:sinx = \dfrac{ \sqrt{3} }{2}  \: when \: x \:  =  \: \dfrac{\pi}{3}}

So,

\bf\implies \:x = \pi + \: \dfrac{\pi}{3} \:  \: or \:  \: 2\pi - \: \dfrac{\pi}{3}

\bf\implies \:x =\: \dfrac{4\pi}{3} \:  \: or \:  \: \dfrac{5\pi}{3}

Hence,

 \purple{ \sf \: Principal \: solution \: of \:  \sqrt{3}cosecx + 2 = 0 \: is \: \dfrac{4\pi}{3} \: or \: \dfrac{5\pi}{3}}

Additional Information :-

\begin{gathered}\boxed{\begin{array}{c|c} \sf T- {eq}^{n}  & \sf General \:  {sol}^{n}  \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2} \\ \\ \sf tanx = 0 & \sf x = n\pi\\ \\ \sf sinx =siny & \sf x = n\pi +  {( - 1)}^{n}\pi \\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\\ \\ \sf tanx = tany & \sf x = n\pi + y \end{array}} \\ \end{gathered}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \bull \:  \:  \: \pink{\bf \: where \: n \:  \in \: integer}

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