18. $12sin^{2}x + 2cos x - 10 = 0$

Explanation:

Step1: Use the identity $\sin^{2}x = 1-\cos^{2}x$

Substitute $\sin^{2}x$ in the equation $12\sin^{2}x + 2\cos x-10 = 0$. We get $12(1 - \cos^{2}x)+2\cos x - 10=0$. Expand the left - hand side: $12-12\cos^{2}x + 2\cos x-10 = 0$. Simplify to $-12\cos^{2}x+2\cos x + 2 = 0$. Multiply through by - 1 to get $12\cos^{2}x-2\cos x - 2 = 0$. Divide by 2: $6\cos^{2}x-\cos x - 1 = 0$.

Step2: Let $t = \cos x$

The equation becomes a quadratic equation $6t^{2}-t - 1=0$. For a quadratic equation $at^{2}+bt + c = 0$ (here $a = 6$, $b=-1$, $c = - 1$), the quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-1)^{2}-4\times6\times(-1)=1 + 24 = 25$. Then $t=\frac{1\pm\sqrt{25}}{2\times6}=\frac{1\pm5}{12}$. We have two solutions for $t$: $t_1=\frac{1 + 5}{12}=\frac{6}{12}=\frac{1}{2}$ and $t_2=\frac{1-5}{12}=\frac{-4}{12}=-\frac{1}{3}$.

Step3: Substitute back $t=\cos x$

When $\cos x=\frac{1}{2}$, $x = 2k\pi\pm\frac{\pi}{3},k\in\mathbb{Z}$. When $\cos x=-\frac{1}{3}$, $x = 2k\pi\pm\arccos(-\frac{1}{3}),k\in\mathbb{Z}$.

Answer:

$x = 2k\pi\pm\frac{\pi}{3},k\in\mathbb{Z}$ and $x = 2k\pi\pm\arccos(-\frac{1}{3}),k\in\mathbb{Z}$