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September 28th, 2016, 12:56 AM   #1
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real values of a

if $\sin x+\cos x= y^2-y+a$ the equation has no real values of $x$ for any real values of $y,$Then values of $a$
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September 28th, 2016, 07:39 AM   #2
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show some work
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September 28th, 2016, 02:42 PM   #3
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$\displaystyle a\gt\sqrt2+\frac14$
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September 29th, 2016, 01:05 AM   #4
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Using greg 1313 hint

$-\sqrt{2} \leq \sin x+\cos x \leq \sqrt{2}\;\forall x\in \mathbb{R}$

So $\displaystyle \underbrace{\sin x+\cos x}_{\bf{\leq \sqrt{2}}} = y^2-y+\frac{1}{4}+a-\frac{1}{4} = \left(y-\frac{1}{2}\right)^2+a-\frac{1}{4}$

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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