Question

Solve the following :

$\bf \: if \: \: \: \rm \: r \: sin \theta = 3 \: \: and \: \: \: r = 4(1 + sin\theta) \\ \sf \: then \: \: (r,\theta) \: \equiv \: ? \\ \\ \rm \: where \: \: \: r > 0 \: \: \: and \: \: - \pi \leqslant \theta \leqslant \pi$
​

Answer 1

Step-by-step explanation:

here you go if you are satisfied thank me

Answer 2

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

Given that,

$\rm :\longmapsto\:rsin\theta = 3 - - - - (1)$

and

$\rm :\longmapsto\:r = 4(1 + sin\theta ) - - - (2)$

So, from equation (1),

$\rm :\longmapsto\:sin\theta = \dfrac{3}{r} - - - - (3)$

Substituting the value from equation (3) in equation (2), we get

$\rm :\longmapsto\:r = 4\bigg[1 + \dfrac{3}{r} \bigg]$

$\rm :\longmapsto\:r = 4\bigg[ \dfrac{r + 3}{r} \bigg]$

$\rm :\longmapsto\: {r}^{2} = 4r + 12$

$\rm :\longmapsto\: {r}^{2} - 4r - 12 = 0$

$\rm :\longmapsto\: {r}^{2} - 6r + 4r - 12 = 0$

$\rm :\longmapsto\:r(r - 6) + 2(r - 6) = 0$

$\rm :\longmapsto\:(r - 6)(r + 2) = 0$

$\bf\implies \:r = 6 \: \: or \: \: r = - \: 2 \: \: \: \{r > 0 \}$

Now, On substituting the value of r, in equation (1), we get

$\rm :\longmapsto\:6sin\theta = 3$

$\rm \implies\:sin\theta = \dfrac{1}{2}$

$\rm \implies\:sin\theta = sin \dfrac{\pi}{6}$

$\red{\bf \implies\:\theta \: = \: n\pi + {( - 1)}^{n}\dfrac{\pi}{6} \: \: \: \forall \: n \in \: Z }$

$\purple{\bf :\longmapsto\:As \: it \: is \: given \: that \: - \pi \leqslant \theta \leqslant \pi}$

$\bf\implies \:\theta \: = \: \dfrac{\pi}{6}, \: \: \dfrac{5\pi}{6}$

Thus,

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:(r,\theta ) = \begin{cases} &\sf{\bigg(6, \: \dfrac{\pi}{6} \bigg)} \\ \\ &\sf{\bigg(6, \: \dfrac{5\pi}{6} \bigg)} \end{cases}\end{gathered}\end{gathered}}$

Additional Information :-

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