Question

Find the integral values of k for which the system of equations;
$\begin{aligned} & arc(\cos x)+{{(arc\,\,\sin y)}^{2}}=\dfrac{k{{\pi }^{2}}}{4} \\\ & {{(arc\,\,\sin y)}^{2}}.\,arc(\cos x)=\dfrac{{{\pi }^{4}}}{16} \\\ \end{aligned}$
Possess the solution and also find the solutions.

Answer 1

We can assume $arc(\cos x)$ and $arc(\sin y)$ as m and n. We rearrange the given equation in the form of m and n. Then we get two quadratic equations in the form of m and n. We solve those equations to get the values of m and n. Now, solve the equations $arc(\cos x)=m$ and $arc(\sin y)=n$ to get the values of $x$ and $y$.

Complete step-by-step answer :
Here, $arc(\cos x)={{\cos }^{-1}}x$ and $arc(\sin y)={{\sin }^{-1}}y$.
We assume $arc(\cos x)$ and $arc(\sin y)$ as m and n.
So, $arc(\cos x)=m$ and $arc(\sin y)=n$.
It implies $\cos m=x$ and $\sin n=y$.
We can easily change the ordered pair $(m,n)$ to $(x,y)$.
Now, the given equations become
$\begin{aligned} & m+{{(n)}^{2}}=\dfrac{k{{\pi }^{2}}}{4} \\\ & {{(n)}^{2}}.\,m=\dfrac{{{\pi }^{4}}}{16} \\\ \end{aligned}$
Now, we solve them to get the values of m and n.
$\begin{aligned} & m+{{(n)}^{2}}=\dfrac{k{{\pi }^{2}}}{4} \\\ & \Rightarrow {{n}^{2}}=\dfrac{k{{\pi }^{2}}}{4}-m \\\ \end{aligned}$
We place the value of ${{n}^{2}}$ in the second equation.
$\begin{aligned} & {{(n)}^{2}}.\,m=\dfrac{{{\pi }^{4}}}{16} \\\ & \Rightarrow (\dfrac{k{{\pi }^{2}}}{4}-m).\,m=\dfrac{{{\pi }^{4}}}{16} \\\ & \Rightarrow 4mk{{\pi }^{2}}-16{{m}^{2}}={{\pi }^{4}} \\\ & \Rightarrow 16{{m}^{2}}-4mk{{\pi }^{2}}+{{\pi }^{4}}=0 \\\ \end{aligned}$
The equation becomes a quadratic equation of m.
The discriminant of the equation will be non-negative. $\Rightarrow D\ge 0$.
So, for the inequation $16{{m}^{2}}-4mk{{\pi }^{2}}+{{\pi }^{4}}=0$
$\begin{aligned} & {{\left( -4k{{\pi }^{2}} \right)}^{2}}-4\times 16\times {{\pi }^{4}}\ge 0 \\\ & \Rightarrow {{\pi }^{4}}({{k}^{2}}-4)\ge 0 \\\ \end{aligned}$
From the inequation we try to find the values of k.
$\begin{aligned} & {{\pi }^{4}}({{k}^{2}}-4)\ge 0 \\\ & \Rightarrow ({{k}^{2}}-4)\ge 0 \\\ & \Rightarrow {{k}^{2}}\ge 4 \\\ \end{aligned}$
Solution of the inequation will be $k\le -2$ or $k\ge 2$.
Now, we take one given equation with the unknown k.
$arc(\cos x)+{{(arc\,\,\sin y)}^{2}}=\dfrac{k{{\pi }^{2}}}{4}$
Here, the principal solution of $arc(\cos x)$ lies in $\left[ 0,\pi \right]$ and $arc(\sin y)$ lies in $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.
So, $arc(\cos x)+{{(arc\,\,\sin y)}^{2}}$ will always be a positive value which means $\dfrac{k{{\pi }^{2}}}{4}$will be a positive value.
It means k can only take positive values.
So, our solutions $k\le -2$ or $k\ge 2$ will automatically omit the negative value, means the only solutions will be $k\ge 2$.
Now, we try to find the maximum value of $arc(\cos x)+{{(arc\,\,\sin y)}^{2}}=\dfrac{k{{\pi }^{2}}}{4}$.
We know that for $arc(\cos x)$ and $arc(\sin y)$ the maximum values are $2\pi$ and $\dfrac{\pi }{2}$.
So, putting maximum values of $arc(\cos x)+{{(arc\,\,\sin y)}^{2}}$
$\begin{aligned} & arc(\cos x)+{{(arc\,\,\sin y)}^{2}}=\dfrac{k{{\pi }^{2}}}{4} \\\ & \Rightarrow 2\pi +{{(\dfrac{\pi }{2})}^{2}}=\dfrac{k{{\pi }^{2}}}{4} \\\ & \Rightarrow \dfrac{k{{\pi }^{2}}}{4}-\dfrac{{{\pi }^{2}}}{4}=2\pi \\\ \end{aligned}$
Now, we solve for k.

& \dfrac{k{{\pi }^{2}}}{4}-\dfrac{{{\pi }^{2}}}{4}=2\pi \\\ & \Rightarrow \dfrac{(k-1){{\pi }^{2}}}{4}=2\pi \\\ & \Rightarrow (k-1)\pi =8 \\\ & \Rightarrow (k-1)=\dfrac{8}{\pi } \\\ & \Rightarrow k=\dfrac{8}{\pi }+1 \\\ \end{aligned}$$ k only takes integral values where $$k=\dfrac{8}{\pi }+1=3.54(approx.)$$ For integral value solutions we get $k=2,3[\because k\ge 2]$ Now, we put the values of k one by one. For $k=2$ the given equation becomes $\begin{aligned} & m+{{(n)}^{2}}=\dfrac{{{\pi }^{2}}}{2} \\\ & {{(n)}^{2}}.\,m=\dfrac{{{\pi }^{4}}}{16} \\\ \end{aligned}$ We solve the equations by replacing the value ${{(n)}^{2}}$ of we get $\begin{aligned} & m+{{(n)}^{2}}=\dfrac{{{\pi }^{2}}}{2} \\\ & \Rightarrow m+\dfrac{{{\pi }^{4}}}{16m}=\dfrac{{{\pi }^{2}}}{2}[\because {{(n)}^{2}}=\dfrac{{{\pi }^{4}}}{16m}] \\\ \end{aligned}$ We solve it by turning into a square form $\begin{aligned} & m+\dfrac{{{\pi }^{4}}}{16m}=\dfrac{{{\pi }^{2}}}{2} \\\ & \Rightarrow 16{{m}^{2}}-8m{{\pi }^{2}}+{{\pi }^{4}}=0 \\\ & \Rightarrow {{\left( 4m-{{\pi }^{2}} \right)}^{2}}=0 \\\ \end{aligned}$ We get the value of m as $$\begin{aligned} & 4m={{\pi }^{2}} \\\ & \Rightarrow m=\dfrac{{{\pi }^{2}}}{4} \\\ \end{aligned}$$ So, now putting value of m we get value of n. $\begin{aligned} & {{(n)}^{2}}.\,m=\dfrac{{{\pi }^{4}}}{16} \\\ & \Rightarrow {{(n)}^{2}}.\,\left( \dfrac{{{\pi }^{2}}}{4} \right)=\dfrac{{{\pi }^{4}}}{16} \\\ & \Rightarrow n=\pm \dfrac{\pi }{2} \\\ \end{aligned}$ So, the final solutions be for $k=2$ the ordered pair be $\left( \dfrac{{{\pi }^{2}}}{4},\pm \dfrac{\pi }{2} \right)$. For changing from $\left( m,n \right)$ we get $\left( x,y \right)$. For $k=2$ the ordered pair be $\left( \cos \left( \dfrac{{{\pi }^{2}}}{4} \right),\sin \left( \pm \dfrac{\pi }{2} \right) \right)=\left( \cos \left( \dfrac{{{\pi }^{2}}}{4} \right),\pm 1 \right)$. For $k=3$ the given equation becomes $\begin{aligned} & m+{{(n)}^{2}}=\dfrac{3{{\pi }^{2}}}{4} \\\ & {{(n)}^{2}}.\,m=\dfrac{{{\pi }^{4}}}{16} \\\ \end{aligned}$ So, solving the equations by replacing the value ${{(n)}^{2}}$ of we get $\begin{aligned} & m+{{(n)}^{2}}=\dfrac{3{{\pi }^{2}}}{4} \\\ & \Rightarrow m+\dfrac{{{\pi }^{4}}}{16m}=\dfrac{3{{\pi }^{2}}}{4}[\because {{(n)}^{2}}=\dfrac{{{\pi }^{4}}}{16m}] \\\ \end{aligned}$ We solve it by quadratic solving method. $\begin{aligned} & m+\dfrac{{{\pi }^{4}}}{16m}=\dfrac{3{{\pi }^{2}}}{4} \\\ & \Rightarrow 16{{m}^{2}}-12m{{\pi }^{2}}+{{\pi }^{4}}=0 \\\ \end{aligned}$ We get the value of m as $$\begin{aligned} & 16{{m}^{2}}-12m{{\pi }^{2}}+{{\pi }^{4}}=0 \\\ & \Rightarrow m=\dfrac{12{{\pi }^{2}}\pm \sqrt{144{{\pi }^{4}}-64{{\pi }^{4}}}}{32} \\\ & \Rightarrow m=\dfrac{3{{\pi }^{2}}\pm {{\pi }^{2}}\sqrt{9-4}}{8} \\\ & \Rightarrow m=\dfrac{{{\pi }^{2}}(3\pm \sqrt{5})}{8} \\\ \end{aligned}$$ So, now putting value of m we get value of n. $$\begin{aligned} & {{(n)}^{2}}.\,m=\dfrac{{{\pi }^{4}}}{16} \\\ & \Rightarrow {{(n)}^{2}}.\,\left( \dfrac{{{\pi }^{2}}(3\pm \sqrt{5})}{8} \right)=\dfrac{{{\pi }^{4}}}{16} \\\ & \Rightarrow {{(n)}^{2}}=\dfrac{{{\pi }^{2}}}{2(3\pm \sqrt{5})} \\\ & \Rightarrow {{(n)}^{2}}=\dfrac{\pi }{6\pm 2\sqrt{5}}={{\left[ \dfrac{\pi }{(\sqrt{5}+1)} \right]}^{2}} \\\ & \Rightarrow n=\pm \dfrac{\pi }{(\sqrt{5}+1)} \\\ \end{aligned}$$ So, the final solutions be for $k=3$ the ordered pair be $\left( \dfrac{{{\pi }^{2}}(3\pm \sqrt{5})}{8},\pm \dfrac{\pi }{(\sqrt{5}+1)} \right)$. For changing from $\left( m,n \right)$ we get $\left( x,y \right)$. For $k=3$ the ordered pair be $\left( \cos \dfrac{{{\pi }^{2}}(3\pm \sqrt{5})}{8},\sin \left( \pm \dfrac{\pi }{(\sqrt{5}+1)} \right) \right)$. In this way we get two solutions in total. For $k=2$ the ordered pair be $\left( \cos \left( \dfrac{{{\pi }^{2}}}{4} \right),\pm 1 \right)$. For $k=3$ the ordered pair be $\left( \cos \dfrac{{{\pi }^{2}}(3\pm \sqrt{5})}{8},\sin \left( \pm \dfrac{\pi }{(\sqrt{5}+1)} \right) \right)$. **Note** : At the time of solving the solution we can take $arc(\cos x)$ and $arc(\sin y)$ as their inverse form, but that only will make the solution more tough to understand. So, the main focus should be to minimize the big equations into smaller ones. Also, getting one solution for the equation also solves the problem.