Question

find the principal solution of cosecx=2 (with picture or solution please )​

Answer 1

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

The given Trigonometric equation is

$\red{\rm :\longmapsto\:cosecx = 2 \: }$

can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{sinx} = 2$

$\rm :\longmapsto\:sinx = \dfrac{1}{2}$

$\rm :\longmapsto\:sinx =sin\bigg[ \dfrac{\pi}{6} \bigg]$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: sinx = sin(\pi - x) \: }}}$

So, using this, above can be rewritten as

$\rm :\longmapsto\:sinx = sin\bigg[\dfrac{\pi}{6} \bigg] \: \: or \: \: sin\bigg[\pi - \dfrac{\pi}{6} \bigg]$

$\rm :\longmapsto\:sinx = sin\bigg[\dfrac{\pi}{6} \bigg] \: \: or \: \: sin\bigg[\dfrac{6\pi - \pi}{6} \bigg]$

$\rm :\longmapsto\:sinx = sin\bigg[\dfrac{\pi}{6} \bigg] \: \: or \: \: sin\bigg[\dfrac{5\pi}{6} \bigg]$

$\bf\implies \:x = \dfrac{\pi}{6} \: \: or \: \: \dfrac{5\pi}{6}$

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More to know :-

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