Question

Solve the following pair of equations for x and y

${4}^{sinx} + {3}^{secy} = 11 \\ \\ 5. {16}^{sinx} - 2. {3}^{secy} = 2$

​

Answer 1

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

Given pair of equations are

$\rm \: {4}^{sinx} + {3}^{secy} = 11 - - - (1)$

and

$\rm \: 5. {16}^{sinx} - 2. {3}^{secy} = 2$

can be rewritten as

$\rm \: 5. {( {4}^{2}) }^{sinx} - 2. {3}^{secy} = 2$

can be further rewritten as

$\rm \: 5. {( {4}^{sinx}) }^{2} - 2. {3}^{secy} = 2 - - - - (2)$

Let assume that

$\rm \: {4}^{sinx} = u \: \: and \: \: {3}^{secy} = v - - - - (3)\bigg[\rm\implies \:u,v > 0\bigg]$

So, equation (1) and (2), can be rewritten as

$\rm \: u + v = 11 - - - - (4)$

and

$\rm \: {5u}^{2} - 2v = 2$

On substituting the value of v from equation (4), we get

$\rm \: {5u}^{2} - 2(11 - u) = 2$

$\rm \: {5u}^{2} - 22 + 2u = 2$

$\rm \: {5u}^{2}+ 2u - 24= 0$

$\rm \: {5u}^{2}+ 12u - 10u - 24= 0$

$\rm \: u(5u + 12) - 2(5u + 12) = 0$

$\rm \: (5u + 12)(u - 2) = 0$

$\rm\implies \:u \: = \: 2 \: \: \: or \: \: u = - \dfrac{5}{12} \{rejected \: as \: u > 0 \}$

On substituting u = 2 in equation (4), we get

$\rm \: 2 + v = 11$

$\rm\implies \:v = 9$

So, we have

$\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\bf{u \: = \: 2} \\ \\ &\bf{v \: = \: 9} \end{cases}\end{gathered}\end{gathered}$

Now, Consider

$\rm \: u = 2$

$\rm \: {4}^{sinx} = 2$

$\rm \: {2}^{2sinx} = {2}^{1}$

$\rm\implies \:2sinx = 1$

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

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

We know,

$\boxed{\tt{ \: sinx = siny \: \: \rm\implies \: \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z}}$

So, using this result, we get

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

Now, Consider

$\rm \: v = 9$

$\rm \: {3}^{secy} = 9$

$\rm \: {3}^{secy} = {3}^{2}$

$\rm\implies \:secy = 2$

$\rm\implies \:cosy = \dfrac{1}{2}$

$\rm\implies \:cosy =cos \dfrac{\pi}{3}$

We know,

$\boxed{\tt{ \: cosx = cosy \: \: \rm\implies \: \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z \: }}$

So, using this result, we get

$\rm\implies \:y = 2p\pi \pm \: \dfrac{\pi}{3}\: \forall \: p \in \: Z$

So, Solution set of simultaneous equations are

$\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\bf{x = n\pi + {( - 1)}^{n}\dfrac{\pi}{6}\: \forall \: n \in \: Z} \\ \\ &\bf{y = 2p\pi \pm \: \dfrac{\pi}{3}\: \forall \: p \in \: Z} \end{cases}\end{gathered}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

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