Find number of solutions in the interval $[0,2\pi]$ of the equation - $$\csc x + \sec x = 2 \sqrt{2}.$$

$⇒ \dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2$

$⇒ \dfrac{\sin x +\cos x}{\sin x \cos x} = 2 \sqrt 2 ⇒ (\sin x+\cos x)^2=8\sin^2x\cos^2x⇒1+2\sin x \cos x=8\sin^2x \cos^2x$

$⇒\sin x \cos x=-1/4 ,\sin x \cos x=1/2⇒2\sin x \cos x=-1/2 , 2\sin x\cos x=1$

$⇒\sin2x =-1/2 , \sin2x=1.$ But if we check manually , $\sin2x=-1/2$ is not a solution.

$⇒\sin 2x =1⇒2x=(-1)^{n}\dfrac{\pi}{2}+n\pi⇒x=(-1)^{n}\dfrac{\pi}{4}+\dfrac{n\pi}{2}⇒$ For $[0,2\pi], \boxed{x=\dfrac{\pi}{4},\dfrac{5 \pi}{4}}$ are solutions.

But if we check the graph neither are the solutions . But $\pi /4 $ is a solution but $5\pi /4 $ is not a solution if we plug values manually. enter image description here

Can someone point out the mistake? Thanks

$\text{References:}$ Link for above graph : https://www.desmos.com/calculator/eaakqvd9y5

  • 1
    $\begingroup$ When you square the equation in the second line, you actually solve the equations $\sin x+\cos x=\pm \sqrt{2} \sin 2 x$, where $\frac{5\pi}{4}$ is exactly the root of $\sin x+\cos x=-\sqrt{2} \sin 2 x$. Therefore check answers when you solve equation by squaring. $\endgroup$
    – Lai
    Commented Sep 7, 2022 at 9:40

2 Answers 2


As noticed in the comments, the issue is that by squaring we can add solutions, as for the following trivial example

$$x=-1 \implies x^2 =1 \implies x=1 \lor x=-1$$

In this case to avoid this issue we can proceed as follows (since $\sin x\cos x\neq 0$):

$$\dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2 \iff \cos x + \sin x = 2 \sqrt2 \sin x \cos x$$


$$2 \sin x \cos x =(\cos x + \sin x)^2-1$$

and thus

$$\sqrt 2(\cos x + \sin x)^2-(\cos x + \sin x)-\sqrt 2=0$$

from wich we obtain

$$\cos x + \sin x=\sqrt 2 \:\lor \: \cos x + \sin x=-\frac{\sqrt 2}2$$

then we can use that $\cos x + \sin x= \sqrt 2 \sin \left(x+\frac \pi 4\right)$ to obtain the result.

As an alternative from here

$$\cos x + \sin x = 2 \sqrt2 \sin x \cos x$$

we have that

$$\sqrt 2 \sin \left(x+\frac \pi 4\right)= \sqrt 2 \sin (2x)$$

$$ \sin \left(x+\frac \pi 4\right)= \sin (2x)$$

which leads to

$$x+\frac \pi 4 = 2x + 2k\pi \: \lor \: x+\frac \pi 4 = \pi -2x + 2k\pi$$

that is $x=\frac \pi 4 +\frac23k\pi$.

Here is a graph with a visualization for the solutions

enter image description here

  • $\begingroup$ So what's the correct final answer? $\endgroup$
    – Aleph
    Commented Sep 7, 2022 at 11:50
  • $\begingroup$ Also why isnt $(\pi/4)$ a solution in the graph? $\endgroup$
    – Aleph
    Commented Sep 7, 2022 at 11:50
  • $\begingroup$ @Aqua As you can easily check, from the last two we obtain three solutions: $x=\frac \pi 4 + \frac 23 k\pi$ $\endgroup$
    – user
    Commented Sep 7, 2022 at 11:52


Instead of squaring $$\dfrac {\sin x + \cos x}{\sin x \cos x} = 2\sqrt{2}$$ clear fractions to get $$\sin x + \cos x = 2\sqrt{2}\sin x \cos x$$ and note that you have a double angle identity hidden on the RHS. Then you can use another identity to get the LHS and find the relationship between the two.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .