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 number of values of $y\in [0,6\pi ]$ is

• A. $5$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is D $6$

$x{\cos }^{3}y+3x\cos y{\sin }^{2}y=14$ ........ (1)

$x{\sin }^{3}y+3x{\cos }^{2}y\sin y=13$ ........ (2)

Adding (1) and (2)

$x({\cos }^{3}y+{\sin }^{3}y+3\cos y{\sin }^{2}y+3{\cos }^{2}y\sin y)=27$

$\Rightarrow x{(\cos y+\sin y)}^3=27$

Subtracting (2) from (1), we get

$x({\cos }^{3}y-{\sin }^{3}y+3\cos y{\sin }^{2}y-3{\cos }^{2}y\sin y)=1$

$\Rightarrow x{(\cos y-\sin y)}^3=1$

$\Rightarrow \frac{\cos y+\sin y}{\cos y-\sin y}=\frac{3}{1}$

apply componendo and dividendo

$\Rightarrow \frac{\cos y+\sin y+\cos y-\sin y}{\cos y+\sin y-\cos y+\sin y}=\frac{3+1}{3-1}$

$\Rightarrow \frac{2\cos y}{2\sin y}=\frac{4}{2}$

$\Rightarrow \frac{\cos y}{\sin y}=\frac{2}{1}$

$\Rightarrow \tan y=\frac{2}{1}$

$\Rightarrow {\tan }^{-1}y=\frac{1}{2}\Rightarrow y=n\pi +{\tan }^{-1}\frac{1}{2}$

but $yϵ[0,6\pi ]$ so $y$ can only have $6$ values i.e., $n=0$ to $5$