September 28th, 2016, 12:56 AM  #1 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16  real values of a
if $\sin x+\cos x= y^2y+a$ the equation has no real values of $x$ for any real values of $y,$Then values of $a$

September 28th, 2016, 07:39 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 
show some work

September 28th, 2016, 02:42 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 
$\displaystyle a\gt\sqrt2+\frac14$

September 29th, 2016, 01:05 AM  #4 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16 
Using greg 1313 hint $\sqrt{2} \leq \sin x+\cos x \leq \sqrt{2}\;\forall x\in \mathbb{R}$ So $\displaystyle \underbrace{\sin x+\cos x}_{\bf{\leq \sqrt{2}}} = y^2y+\frac{1}{4}+a\frac{1}{4} = \left(y\frac{1}{2}\right)^2+a\frac{1}{4}$ So $\displaystyle a\frac{1}{4}>\sqrt{2}$ for no real solution. So we get $\displaystyle a>\sqrt{2}+\frac{1}{4}$ 

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