Does there exist \alpha \in (0,\dfrac{\pi}{2}) such that the functions x^2-4x\cos \alpha +2 = 0 and x^2-4x\sin \alpha -2=0 in terms of x have an equal real root? If there exists, find the value of \alpha . If not, explain why.
x^2-4x\cos \alpha +2 = 0
x=\dfrac{4\cos \alpha \pm\sqrt{16\cos^2\alpha -8 } }{4}
x^2-4x\sin \alpha -2=0
x=\dfrac{4\sin \alpha \pm\sqrt{16\sin^2\alpha +8 } }{-4}
Since the radicand of the radical of (1) is larger or equal to zero,
16\cos^2 \alpha -8 \geq 0
\cos^2 \alpha \geq \dfrac{1}{2}
\alpha \leq \dfrac{\pi}{6}
Therefore, cancel the negative case of (1) and (2) and yield the following equation.
\cos \alpha + \dfrac{1}{2} \sqrt{4\cos^2 \alpha -2 } = -\sin \alpha + \dfrac{1}{2}\sqrt{4\sin^2 \alpha +2 }
2(\sin \alpha +\cos \alpha ) = \sqrt{4\sin^2 \alpha +2 }- \sqrt{4\cos^2 \alpha -2 }
4(1+2\sin \alpha \cos \alpha )
= 4\sin ^2 \alpha +2 + 4\cos ^2 \alpha -2 - 4 \sqrt{ 2\sin^2 \alpha+1}\sqrt{2 \cos^2 \alpha-1}
2\sin \alpha \cos \alpha = \sqrt{ 2\sin^2 \alpha+1}\sqrt{2 \cos^2 \alpha-1}
4\sin^2 \alpha \cos^2 \alpha = (2\sin^2 \alpha+1)(2 \cos^2 \alpha-1)
= 4\sin^2 \alpha \cos^2 \alpha - 2\sin^2 \alpha + 2 \cos^2 \alpha -1
Therefore,
\cos^2 \alpha - \sin^2 \alpha = \dfrac{1}{2}
\cos 2\alpha = \dfrac{1}{2}
2\alpha = \dfrac{\pi}{3}
\alpha = \dfrac{\pi}{6}