Trigonometric equations[1 MORE]

Hi,

Could someone check this for me please.

Solve

$\displaystyle 6 cos^2{\theta} + sin{\theta} = 4\, for\, 0 \le x \le {\pi}$

$\displaystyle 6(1 - sin^2{\theta}) + sin{\theta} = 4$

$\displaystyle -6sin^2{\theta} + sin{\theta} + 2 = 0$

$\displaystyle \text{Let}$ $\displaystyle sin{\theta} = a, |a| \le 1$

$\displaystyle 6a^2 - a - 2 = 0$

$\displaystyle D = 1 - 4(6)(-2) = 49$

$\displaystyle a_1 = \frac{-(-1) + \sqrt49}{(2)(6)}$

$\displaystyle = \frac{8}{12} $

$\displaystyle = \frac{2}{3}$

$\displaystyle a_2 = \frac{-(-1) - \sqrt49}{(2)(6)}$

$\displaystyle = \frac{-1}{2}$

$\displaystyle sin{\theta}_1 = \frac{2}{3}$

$\displaystyle sin{\theta}_2 = \frac{-1}{2}$

$\displaystyle 0 \le x \le {\pi}$

$\displaystyle \Rightarrow$$\displaystyle sin{\theta} > 0$

$\displaystyle \Rightarrow$ $\displaystyle sin{\theta} = \frac{2}{3}$

$\displaystyle {\theta} = (-1)^n $

$\displaystyle arcsin$ $\displaystyle \frac{2}{3}$ $\displaystyle + {\pi}n$

$\displaystyle \text{Solution:}\,$$\displaystyle {\theta}$$\displaystyle = arcsin \frac{2}{3}, n = 0$

$\displaystyle {\theta} = {\pi} - arcsin \frac{2}{3}, n = 1$