Question

Consider the system of equations
$x{\cos }^{3}y+3x\cos y{\sin }^{2}y=14$
$x{\sin }^{3}y+3x{\cos }^{2}y\sin y=13$

The value(s) of $x$ is/are

• A. $\pm 5\sqrt{5}$
• B. $\pm \sqrt{5}$
• C. $\pm \frac{1}{\sqrt{5}}$
• D. None of these

Answer 1

The correct option is A $\pm 5\sqrt{5}$

$x{\cos }^{3}y+3x\cos y{\sin }^{2}y=14\cdots \left(1\right)$
$x{\sin }^{3}y+3x{\cos }^{2}y\sin y=13\cdots \left(2\right)$
Adding Eqs. $\left(1\right)$ and $\left(2\right)$
$x({\cos }^{3}y+3{\cos }^{2}y\sin y+3\cos y{\sin }^{2}y+{\sin }^{3}y)=27$
$\Rightarrow x(\cos y+\sin y{)}^3=27$
$\Rightarrow x^{1/3}(\cos y+\sin y)=3\cdots \left(3\right)$

Subtracting Eq $\left(2\right)$ from Eq $\left(1\right)$, we get
$x({\cos }^{3}y+3\cos y{\sin }^{2}y-3{\cos }^{2}y\sin y-{\sin }^{3}y)=1$
$\Rightarrow x(\cos y-\sin y{)}^3=1$
$\Rightarrow x^{1/3}(\cos y-\sin y)=1\cdots \left(4\right)$

Dividing Eq $\left(3\right)$ by Eq $\left(4\right)$, we get
$\cos y+\sin y=3\cos y-3\sin y$
$\Rightarrow \tan y=\frac{1}{2}$

Case $\text{I}:$
$\sin y=\frac{1}{\sqrt{5}},\cos y=\frac{2}{\sqrt{5}}$
Then $y$ lies in the $1$st quadrant
$\therefore y=2n\pi +\alpha$, where $0<\alpha <\frac{\pi }{2}$ and $\sin \alpha =\frac{1}{\sqrt{5}}$
From Eq.$\left(3\right)$, we get
$x^{1/3}\left(\frac{3}{\sqrt{5}}\right)=3$
$\Rightarrow x=5\sqrt{5}$

Case $\text{II}:$
$\sin y=\frac{-1}{\sqrt{5}},\cos y=\frac{-2}{\sqrt{5}}$
Then $y$ lies in the $3$rd quadrant
$\therefore y=2n\pi +(\pi +\alpha )$, where $0<\alpha <\frac{\pi }{2}$ and $\sin \alpha =\frac{-1}{\sqrt{5}}$
From Eq.$\left(3\right)$, we get
$x^{1/3}\left(\frac{-3}{\sqrt{5}}\right)=3$
$\Rightarrow x=-5\sqrt{5}$