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I want to solve this equation $$\sin\dfrac{(x+1)\pi}{4x^2 -4x + 2} = \cos\dfrac{(x-2)\pi}{4x^2 -4x + 2}.$$ The equation has solutions $x = 1$, $x = \dfrac{1 \pm \sqrt{5}}{2}$ and $x = \dfrac{5\pm \sqrt{5}}{10}$
I think, this is a nonstandard problem. I tried $$\cos\dfrac{(x-2)\pi}{4x^2 -4x + 2} =\sin\left( \dfrac{\pi}{2} - \dfrac{(x-2)\pi}{4x^2 -4x + 2}\right).$$ But I can not solve.
$$\cos\dfrac{(x-2)\pi}{4x^2 -4x + 2}=\sin\dfrac{(x+1)\pi}{4x^2 -4x + 2} =\cos\left( \frac \pi 2-\frac{(x+1)\pi}{4x^2 -4x + 2}\right)$$
So, $$\frac{(x-2)\pi}{4x^2 -4x + 2}=2n\pi\pm \left( \frac \pi 2-\frac{(x+1)\pi}{4x^2 -4x + 2}\right)$$ where $n$ is any integer as $\cos A=\cos B\implies A=2n\pi\pm B$
Multiplying either side by $\frac2\pi,$
$$\frac{x-2}{2x^2-2x+1}=4n\pm \frac{2x^2-3x}{2x^2-2x+1}$$
Taking the '+' sign, $(8n+2)x^2-(8n+4)x+4n+2=0$
for real $x,$ the discriminant $(8n+4)^2-4(8n+2)(4n+2)\ge 0\implies n(2n+1)\le 0$
Now we know if $(x-a)(x-b)\le 0$ where $a\le b\implies a\le x\le b$
Here $a,b$ are $-\frac 12,0$ so, $-\frac 12\le n\le 0\implies n=0 $
So, the quadratic equation of $x$ becomes $x^2-2x+1=0,x=1$
Taking the '-' sign, we can proceed similarly to find
$(4n-1)x^2-(4n-1)x+(2n+1)=0$
and using discriminant property, $(4n-1)^2-4(4n-1)(2n+1)\ge 0\implies (4n-1)(4n+5)\le 0$
$\implies -\frac 54\le n\le \frac 1 4\implies n=0,-1$.
If $n=0,$ the equation reduces to $x^2-x-1=0$
If $n=-1,$ the equation reduces to $5x^2+5x+1=0$
There shall be no other solution of x $\in R$
Your approach is fine. You write $$\sin\left(\frac{(x+1)\pi}{4x^2 -4x + 2}\right) = \cos\left(\frac{(x-2)\pi}{4x^2 -4x + 2}\right)$$ convert to $\sin$ as you suggested $$\sin\left(\frac{(x+1)\pi}{4x^2 -4x + 2}\right) = \sin\left(\frac{\pi}{2} - \frac{(x-2)\pi}{4x^2 -4x + 2}\right)$$ Taking $\arcsin$ over all branches $$\frac{(x+1)\pi}{4x^2 -4x + 2} = k\pi + (-1)^k\left(\frac{\pi}{2} - \frac{(x-2)\pi}{4x^2 -4x + 2}\right)$$ where $k$ is an arbitrary integer. This gives a (messy) quadratic.
We have $\sin a=\sin b$ iff $a-b$ is a multiple of $2pi$ or $a+b$ is an odd multiple of $\pi$. Thus with your conversion fo $\cos$ to $\sin$, you have to find all solutions of $$\frac{(x+1)\pi}{4x^2-4x+2}-\frac\pi2+\frac{(x-2)\pi}{4x^2-4x+2}=2k\pi$$ and of $$\frac{(x+1)\pi}{4x^2-4x+2}+\frac\pi2-\frac{(x-2)\pi}{4x^2-4x+2}=(2k+1)\pi$$ with $k\in\mathbb Z$. Both equations are readily multiplied by $\frac{4x^2-4x+2}\pi$ to produce a quadratic equation (with parameter $k$).
I don't know if this is a step forward or backward, but I was able to solve it like this.
$$\sin[\frac{(x-\frac12)\pi+\frac{3\pi}2}{4x^2-4x+2}]=\cos[\frac{(x-\frac12)\pi-\frac{3\pi}2}{4x^2-4x+2}]$$
Substituting $u=\frac{(x-\frac12)\pi}{4x^2-4x+2}$ and $v=\frac{\frac{3\pi}2}{4x^2-4x+2}$, we get
$$\sin(u+v)=\cos(u-v)$$ $$\sin u\cos v+\sin v\cos u=\cos u\cos v+\sin u\sin v$$ $$\cos u(\cos v-\sin v)-\sin u(\cos v-\sin v)=0$$ $$(\cos u-\sin u)(\cos v-\sin v)=0$$
So we have either $\cos u=\sin u$ or $\cos v=\sin v$. Since sine and cosine can't both be $0$, this can be written as
$$\tan u=1\text{ or }\tan v=1$$ $$u=\frac\pi4+k\pi\text{ or }v=\frac\pi4+k\pi$$
Let's also make one last substitution $y=x-\frac12$. This makes the first equation
$$\frac{\pi y}{4y^2+1}=u$$ $$4uy^2-y\pi+u=0$$
which has real solutions when $\pi^2-16u^2\ge0$, or $-\frac\pi4\le u\le\frac\pi4$. The only valid solution here is $u=\frac\pi4$, which will yield $y=\frac12,x=1$.
Now for the second equation, note that since the denominator is greater than or equal to 1, then $$0<\frac{\frac{3\pi}2}{4y^2+1}\le\frac{3\pi}2$$
so there are solutions where $v=\frac\pi4\text{ or }\frac{5\pi}4$.
