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

Question

I would like to share a trigonometry question here. Wonder is there another way to solve it or not.

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

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

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

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

$(2\sin(x)+1)(\sin(x)+\cos(x)-1)=0$

So, we have $$2\sin(x)+1=0 \space\text{and}\sin(x)+\cos(x)-1=0$$

$2\sin(x)+1=0$

$\sin(x)=-\frac{1}{2}$

$x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n$

$\sin(x)+\cos(x)-1=0$ which answered in my yesterday post $\cos x+\sin x=1$

and get $x=2n\pi,2n\pi+\frac{\pi}{2}$

Combine all the solutions, $$x=2\pi n,x=\frac{7\pi}{6}+2\pi n,x=2\pi n+\frac{\pi}{2}, x=\frac{11\pi}{6}+2\pi n$$

Answer 1

If $\cos A-\sin A=\cos B-\sin B$

As $\cos A-\sin A=\cdots=\sqrt2\cos\left(A+\dfrac\pi4\right),$

$\cos\left(A+\dfrac\pi4\right)=\cos\left(B+\dfrac\pi4\right)$

$\implies A+\dfrac\pi4=2m\pi\pm\left(B+\dfrac\pi4\right)$ where $m$ is any integer

Answer 2

If $\sin2A-\sin2B=\cos2A-\cos2B$

using Prosthaphaeresis Formulas, $$2\sin(A-B)\cos(A+B)=-2\sin(A-B)\sin(A+B)$$

$$2\sin(A-B)[\cos(A+B)+\sin(A+B)]=0$$

$\sin(A-B)=0\implies A-B=n\pi$ where $n$ is any integer

$\cos(A+B)+\sin(A+B)=0\iff\tan(A+B)=-1=\tan\left(-\dfrac\pi4\right)$

$A+B=m\pi-\dfrac\pi4$ where $m$ is any integer