Odpowiedź:
[tex]x\in\left\{-\dfrac{7\pi}{4},\ -\dfrac{5\pi}{4},\ -\dfrac{3\pi}{4},\ -\dfrac{\pi}{4},\ \ \dfrac{\pi}{4},\ \dfrac{3\pi}{4},\ \ \dfrac{5\pi}{4}\right\}[/tex]
Szczegółowe wyjaśnienie:
[tex]\left|\sin\left(\dfrac{\pi+2x}{2}\right)\right|=\sin(\pi-x);\ x\in\left(-2\pi,\ \dfrac{3\pi}{2}\right)\\\\\left|\sin\left(\dfrac{\pi}{2}+\dfrac{2x}{2}\right)\right|=\sin(\pi-x)\\\\\left|\sin\left(\dfrac{\pi}{2}+x\right)\right|=\sin(\pi-x)[/tex]
[tex]\sin\left(\dfrac{\pi}{2}+x\right)=\cos x\\\\\sin(\pi-x)=\sin x[/tex]
Czyli równanie przyjmuje postać:
[tex]|\cos x|=\sin x[/tex]
[tex]|\cos x|=\sin x\iff\cos x=\sin x\ \vee\ \cos x=-\sin x[/tex]
[tex]\cos x=\sin x[/tex] - patrz załącznik
[tex]\huge\boxed{x\in\left\{-\dfrac{7\pi}{4},\ -\dfrac{3\pi}{4},\ \dfrac{\pi}{4},\ \dfrac{5\pi}{4}\right\}}[/tex]
[tex]\cos x=-\sin x\qquad|+\sin x\\\\\sin x+\cos x=0\qquad|^2\\\\(\sin x+\cos x)^2=0\qquad|(a+b)^2=a^2+2ab+b^2\\\\\sin^2x+2\sin x\cos x+\cos^2 x=0\\\\(\sin^2x+\cos^2x)+2\sin x\cos x=0[/tex]
zastosujemy:
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\2\sin\alpha\cos\alpha=\sin2\alpha[/tex]
otrzymujemy:
[tex]1+\sin2x=0\qquad|-1\\\\\sin2x=-1\iff 2x=\dfrac{3}{2}\pi+2k\pi\qquad|:2\\\\x=\dfrac{3\pi}{4}+k\pi\\\\\huge\boxed{x\in\left\{-\dfrac{5\pi}{4},\ -\dfrac{\pi}{4},\ \dfrac{3\pi}{4}\right\}}[/tex]
