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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

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    $\begingroup$ 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

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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

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

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

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

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  • $\begingroup$ 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
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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.

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