Giải thích các bước giải:
$4\sin (x+\dfrac{\pi}{3})\cos (x-\dfrac{\pi}{6})=a^2+\sqrt{3}\sin 2x-\cos 2x$
$\to 4\cos (x-\dfrac{\pi}{6})\cos (x-\dfrac{\pi}{6})=a^2+2(\dfrac{\sqrt{3}}{2}\sin 2x-\dfrac 12\cos 2x)$
$\to 4\cos^2(x-\dfrac{\pi}{6})=a^2+2(\sin 2x\cos\dfrac{\pi}{6}-\cos 2x\sin\dfrac{\pi}{6})$
$\to 4\cos^2(x-\dfrac{\pi}{6})=a^2+2\sin( 2x-\dfrac{\pi}{6})$
$\to 2(2\cos^2(x-\dfrac{\pi}{6})-1)-2\sin( 2x-\dfrac{\pi}{6})=a^2-2$
$\to 2\cos 2(x-\dfrac{\pi}{6})-2\sin( 2x-\dfrac{\pi}{6})=a^2-2$
$\to 2\cos(2x-\dfrac{\pi}{3})-2\sin( 2x-\dfrac{\pi}{6})=a^2-2$
$\to 2(\cos(2x-\dfrac{\pi}{3})-\sin( 2x-\dfrac{\pi}{6}))=a^2-2$
$\to 2(\dfrac{1}{2}.\cos(2x)+\dfrac{\sqrt{3}}{2}\sin(2x)-(-\dfrac{1}2.\cos(2x)+\dfrac{\sqrt{3}}{2}\sin(2x))=a^2-2$
$\to 2\cos(2x)=a^2-2$
$\to -2\le a^2-2\le 2$
$\to 0\le a^2\le 4$
$\to -2\le a\le 2$
$\to$Có 5 giá trị a thỏa mãn đề
