函数 $f(x)=\dfrac{1+\sqrt 3\sin 2x+\cos 2x}{1+\sin x+\sqrt 3\cos x}$ 的值域为_______．
答案 $[-3,1]$．
解析 根据题意，有\[f(x)=\dfrac{1+2\sin\left(2x+\dfrac{\pi}6\right)}{1+2\sin\left(x+\dfrac{\pi}3\right)},\]令 $t=\sin\left(x+\dfrac{\pi}3\right)$，则\[\sin\left(2x+\dfrac{\pi}6\right)=\sin\left(2\left(x+\dfrac{\pi}3\right)-\dfrac {\pi}2\right)=-\cos\left(2\left(x+\dfrac{\pi}3\right)\right)=2t^2-1,\]于是\[\dfrac{1+2\sin\left(2x+\dfrac{\pi}6\right)}{1+2\sin\left(x+\dfrac{\pi}3\right)}=2t-1,\]其中 $-1\leqslant t\leqslant 1$且$t\ne -\dfrac 12$．因此所求值域为 $[-3,-2)\cup(-2,1]$．