trigonometric equation

**jacks** · January 4th 2011, 03:27 AM

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

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

**SammyS** · January 4th 2011, 11:55 AM

Originally Posted by jacks

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

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

What have you tried?

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

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

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

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

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

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

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