# trigonometric equation

• Jan 4th 2011, 03:27 AM
jacks
trigonometric equation
Let $\displaystyle -\frac{\pi}{4}\leq x,y\leq\frac{\pi}{4}$ such that for some positive integer $\displaystyle n ,nx^3+sinx-2010=0$ and

$\displaystyle 4ny^3+sinx.cosy+1005=0,$ then the value of $\displaystyle sin(x+2y)+cos(x+2y)$ is
• Jan 4th 2011, 11:55 AM
SammyS
Re: trigonometric equation (by jacks)
Quote:

Originally Posted by jacks
Let $\displaystyle -\frac{\pi}{4}\leq x,y\leq\frac{\pi}{4}$ such that for some positive integer $\displaystyle n ,nx^3+sinx-2010=0$ and

$\displaystyle 4ny^3+sinx.cosy+1005=0,$ then the value of $\displaystyle sin(x+2y)+cos(x+2y)$ is

What have you tried?

$\displaystyle nx^3+\sin x-2010=0\ \to\ nx^3+\sin x=2010\ \to\ x>0$

For $\displaystyle -\frac{\pi}{4}\leq y\leq\frac{\pi}{4},\ {{1}\over{\sqrt{2}}}\le \cos y \le 1\,.$ Therefore, $\displaystyle y<0\,.$

Solving $\displaystyle n({\pi\over4})^3+\sin ({\pi\over4})-2010=0\,, \text{ for } n \text{ gives }n\approx4147.38\,.$ Thus $\displaystyle n>4157$.

From $\displaystyle n>4157$ that, the second equation gives (approximately) $\displaystyle -{{\pi} \over{8}}\le y<0\,.$

I've only been able to solve this graphically/numerically. All the solutions give:

$\displaystyle \sin(x+2y)+\cos(x+2y)\approx1$