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Find a solution to the equation

$$\frac{1}{\sin^2(2x)}+\tan(x)-\frac{1}{\tan(x)}=2.$$

Assuming $\sin(x) \neq 0$ and $\cos(x) \neq 0$, I simplified the above to

$$4\sin^2(x)-1-4\sin(x)\cos(x)=0,$$

which is equivalent to

$$\cos(2x)+\sin(2x)=\frac{1}{2}.$$

I do not know what to do next. I would be grateful for any hints.

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  • $\begingroup$ It's not going to simplify well, though with $\frac 1{\sin(x)^2}$ instead it simplifies. $\endgroup$
    – zwim
    Nov 13, 2019 at 17:59

4 Answers 4

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We have that by linear combination of sine and cosine

$$\cos(2x)+\sin(2x)=\sqrt 2\sin\left(2x+\frac \pi 4\right)=\frac{1}{2}$$

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  • $\begingroup$ The equality $\cos(2x)+\sin(2x)=\dfrac12$ given by the O.P. I think it is not correct and for the answer $x=\dfrac{\pi}{4}$ your expression gives $1$. Regards. $\endgroup$
    – Piquito
    Nov 13, 2019 at 17:41
  • $\begingroup$ @Piquito Sorry, I didn't check all the works but only the final step! I'll revise that. Bye $\endgroup$
    – user
    Nov 13, 2019 at 17:43
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HINT.- It would be enough to open your eyes wide and realize that $x=\dfrac{\pi}{4}$ (besides of $x=\dfrac{\pi}{4}+k\pi)$ satisfies the equation. In fact, one has for this value $$\frac11+1-0=2$$

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Use $\tan2A=\dfrac{2\tan A}{1-\tan^2A},$

$$\dfrac1{\tan x}-\tan x=\cdots=2\cot2x$$

$$\dfrac1{\sin^22x}=1+\cot^22x$$

On simplification

$$1=\dfrac{2\tan2x}{1-\tan^22x}=\tan4x$$

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As I said in comment I suspect an error in the wording, the equation given really doesn't lead to something pleasant.

Instead with $\dfrac 1{\sin(x)^2}=\dfrac{2}{1-\cos(2x)}=\dfrac{2}{1-\frac{1-t^2}{1+t^2}}=\dfrac{2(1+t^2)}{1+t^2-1+t^2}=\dfrac{1+t^2}{t^2}$

The equation becomes

$\dfrac{1+t^2}{t^2}+t-\frac 1t-2=\dfrac{1+t^2+t^3-t-2t^2}{t^2}=\dfrac{1-t^2+t^3-t}{t^2}=\dfrac{(1-t^2)(1-t)}{t^2}=\dfrac{(1-t)^2(1+t)}{t^2}$

And $\tan(x)=\pm 1$ is found.

With original problem we get to $1-6t^2+t^4+4t^3-4t$

Whose roots are $-1\pm\sqrt{2}\pm\sqrt{4\mp2\sqrt{2}}$ and it is harder to recognize $\dfrac{\pi}{16}$ lines.

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