What is wrong in this solution of $\sec x + \csc x = 2 \sqrt{2}$

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.

Can someone point out the mistake? Thanks

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

• 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.
– Lai
Commented Sep 7, 2022 at 9:40

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 $$\sin x\cos x\neq 0$$):

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

with

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

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

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

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

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

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

Here is a graph with a visualization for the solutions

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

Hint:

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.