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
 A: $$x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}\\
\implies x=\frac{-\frac{2}{\sqrt{\cos t}} \pm\sqrt{\frac{4}{\cos t}
-4(1)\big(\frac{1}{\sin t}-2\sqrt{2}\big)}}{2}\\
=\frac{-1}{\sqrt{\cos t}}\pm \sqrt{\sec {t}-\csc{t} +2\sqrt{2}}$$
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.
A: You face a quadratic equation in $x$
$$x^2+a x+b=0 \qquad \text{with} \qquad a=\frac{2}{\sqrt{\cos (t)}}\quad \text{and} \quad b=\csc (t)-2 \sqrt{2}$$ The discriminant is
$$\Delta=a^2-4 b=4 \left(\sec (t)-\csc (t)+2 \sqrt{2}\right)$$ must be zero to have a double root.
Using the tangent half-angle substitution $t=2 \tan ^{-1}(x)$
$$\sec (t)-\csc (t)+2 \sqrt{2}=-\frac{x^2+1}{2 x}-\frac{2}{x^2-1}+2 \sqrt{2}-1$$ So, what is left is
$$x^4+\left(2-4 \sqrt{2}\right) x^3+\left(2+4 \sqrt{2}\right) x-1=0$$ whih has a double root
$$x_{1,2}=1+\sqrt{2}$$ and what is left is
$$x^2+\left(4-2 \sqrt{2}\right) x+2 \sqrt{2}-3=0$$ which shows the ugly roots
$$x_3=-2+\sqrt{2}-\sqrt{9-6 \sqrt{2}}\qquad \text{and} \qquad x_4=-2+\sqrt{2}-\sqrt{9-6 \sqrt{2}}$$
Use your pocket (or Google) calculator; you will find whole numbers in degrees. Convert to radians and you will obtain the results given by Wolfram Alpha (do not forget the modulo $2\pi$).
A: Note:
$$\sin t\cos \frac{\pi}{4}-\cos t \sin \frac{\pi}{4}=\sin\left(t-\frac{\pi}4\right)$$
So, you must have
$$\sin\left(t-\frac{\pi}{4}\right) =-\sin2t$$
Then you can not simply equate the arguments, because you must remember the period.
Here is how you can continue:
$$\sin\left(t-\frac{\pi}{4}\right)+\sin 2t=0 \stackrel{{\sin A + \sin B=2\sin \frac{A+B}{2}}\cos \frac{A-B}{2}}{\Rightarrow} \\
2\sin{\left(\frac{3t}2-\frac{\pi}{8}\right)}\cos\left(-\frac{t}{2}-\frac{\pi}{8}\right)=0 \Rightarrow \\
\sin\left(\frac{3t}2-\frac{\pi}{8}\right)=0 \quad \text{or} \quad \cos\left(-\frac{t}{2}-\frac{\pi}{8}\right)=0 \Rightarrow \\
\frac{3t}2-\frac{\pi}{8}=\pi n \quad \text{or} \quad -\frac{t}{2}-\frac{\pi}{8}=-\frac{\pi}{2}+\pi n \Rightarrow \\
t=\frac{\pi}{12}+\frac{2\pi n}{3} \quad \text{or} \quad t=\frac{3\pi}4-2\pi n,n\in Z.$$
