Đáp án: $x\in\{\arcsin(\sqrt{\dfrac15})+k2\pi, \pi-\arcsin(\sqrt{\dfrac15})+k2\pi,\arcsin(-\sqrt{\dfrac15})+k2\pi, \pi-\arcsin(-\sqrt{\dfrac15})+k2\pi \}$
Giải thích các bước giải:
Ta có:
$\sin x+\sin^2(\dfrac{x}{2})=\dfrac12$
$\to 2\sin x+2\sin^2(\dfrac{x}{2})=1$
$\to 2\sin x=1-2\sin^2(\dfrac{x}{2})$
$\to 2\sin x=\cos x$
Mà $\sin^2x+\cos^2x=1$
$\to \sin^2x+(2\sin x)^2=1$
$\to 5\sin^2x=1$
$\to \sin^2x=\dfrac15$
$\to \sin x=\sqrt{\dfrac15}\to x=\arcsin(\sqrt{\dfrac15})+k2\pi$ hoặc $x=\pi-\arcsin(\sqrt{\dfrac15})+k2\pi$
Hoặc $ \sin x=-\sqrt{\dfrac15}\to x=\arcsin(-\sqrt{\dfrac15})+k2\pi$ hoặc $x=\pi-\arcsin(-\sqrt{\dfrac15})+k2\pi$
