Find the range of $|a|$ in $ \sec \theta+\csc \theta=a,$ where $\theta \in (0,2\pi)-\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\} $

Question

Find the range of $|a|$ in $ \sec \theta+\csc \theta=a,$ where $\theta \in (0,2\pi)-\{\frac{\pi}{2},\pi,\frac{3\pi}{2}\} $ if the equation has

1) four real roots

2) two real roots

3) no real roots

I have tried converting everything to $\sin$ and using double angle formula and then find range , I have tried squaring both the sides. I was thinking about graphs but how am I suppose to know graph such complex functions?

I am a high school student.

Answer 1

$$a^2=\sec^2t+\csc^2t+2\sec t\csc t=(2\csc2t+1)^2-1$$

Now if $\csc2t>0,2\csc2t+1\ge3,a^2\ge9-1\implies|a|\ge2\sqrt2$

If $\csc2t<0, 2\csc2t+1\le-1,a^2\ge0$ which holds true for all real $a$

So, all real values of $a$ admit solutions

If $\csc2t=-1,a^2=1;$

$2t=2n\pi-\dfrac\pi2\iff t=?$ so, two solutions in $(0<t<2\pi)$

Similarly for $\csc2t=1$

If $\csc2t>1$

$\implies2t$ lies in the first two quadrants and $a^2>8,$

there will be four solutions in $0<2t<4\pi$

What if $\csc2t<1?$

Answer 2

I will do the last case. That is, find bounds on $a$ such that the resulting equation has no roots.

The equation yields $$\sin\theta+\cos\theta=a\sin(2\theta)\implies 2\sqrt 2\cos\left(\theta-\frac π4\right)=a\sin(2\theta).$$

Thus if you take $|a|\ne 2\sqrt 2$ there would definitely not be a solution. You can check the case when $|a| = 2\sqrt 2$ separately to see if there are solutions given the other conditions of the problem. This should tell you either that there are no solutions for all $|a|,$ or on the contrary just for $|a|\ne 2\sqrt 2,$ definitively settling the third part.