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Math Help - trigonometric equation

  1. #1
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    trigonometric equation

    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
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  2. #2
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    Re: trigonometric equation (by jacks)

    Quote Originally Posted by jacks View Post
    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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