<s>[INST] For which value of $t \in \mathbb R$ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$ For which value of $t \in R $ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$
Here $t \neq n\pi, t \neq (2n+1)\frac{\pi}{2}$
Therefore , for the given equation to have exactly one solution  we should have :
$(\frac{2}{\sqrt{\cos t}})^2 -4.(\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \frac{4}{\cos t} - 4 (\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \sin t -\cos t +2 \sqrt{2}\sin t\cos t = 0 $
$\Rightarrow \sqrt{2}( \frac{1}{\sqrt{2}}\sin t - \frac{1}{\sqrt{2}}\cos t) = -2\sqrt{2} \sin t\cos t$
$\Rightarrow \sqrt{2}(\cos(\pi/4)\sin t -\sin(\pi/4)cos t = -\sqrt{2}\sin2t                                  $ [Using $\sin x\cos y -\cos x\sin y = \sin(x-y)$]
$\Rightarrow \sqrt{2}\sin(\frac{\pi}{4}-t) =-\sqrt{2}\sin2t$
$\Rightarrow \sin(\frac{\pi}{4}-t) =-\sin2t                   $                                 [ Using $-\sin x = \sin(-x)$ and comparing R.H.S. with L.H.S. ]
$\Rightarrow  \frac{\pi}{4}-t = -2t $
$\Rightarrow t = - \frac{\pi}{4}$
Is it correct answer, please suggest.. thanks
 [/INST] $$x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}\\
\implies x=\frac{-\frac{2}{\sqrt{\cos t}} \pm\sqrt{\frac{4}{\cos t}
-4(1)\big(\frac{1}{\sin t}-2\sqrt{2}\big)}}{2}\\
=\frac{-1}{\sqrt{\cos t}}\pm \sqrt{\sec {t}-\csc{t} +2\sqrt{2}}$$
Now, the problem is finding what values of $t$ yield
$\quad
\sec t - \csc t + 2\sqrt{2}=0
\quad$
Wolfram Alpha shows infinite $t$-values
here where the equation has only one solution.
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