# Completing the Square Where It Matters

### Solution

$\displaystyle 3\tan^2x+2\tan x+3=3\left(\tan x+\frac{1}{3}\right)^2+\frac{8}{3},\\ \displaystyle \sin^2y+\sin y+1=\left(\sin y+\frac{1}{2}\right)^2+\frac{3}{4}.$

Thus the problem becomes

$\displaystyle \left[3\left(\tan x+\frac{1}{3}\right)^2+\frac{8}{3}\right]\cdot \left[\left(\sin y+\frac{1}{2}\right)^2+\frac{3}{4}\right]=2.$

This reduces to

$\displaystyle 3\left(\tan x+\frac{1}{3}\right)^2\left(\sin y+\frac{1}{2}\right)^2+\frac{9}{4}\left(\tan x+\frac{1}{3}\right)^2+\frac{8}{3}\left(\sin y+\frac{1}{2}\right)^2=0,$

where in the left-hand side all the terms are non-negative, meaning that necessarily

$\displaystyle \sin y=-\frac{1}{2}\;\mbox{and}\;\tan y=-\frac{1}{3}.$

In other words,

$\displaystyle y=(-1)^{k+1}\frac{\pi}{6}+k\pi,\;k\in\mathbb{Z},\\ \displaystyle x=-\arctan\frac{1}{3}+m\pi,\;m\in\mathbb{Z}.$

### Acknowledgment

The problem, with the above solution, has been posted at the CutTheKnotMath facebook page by Dan Sitaru. The problem is from Dan's book "Math Accent".