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
Best Answer
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 $2\pi$).