i) A relação fundamental da trigonometria nos diz que: $sen^2 x + cos^2x = 1$. Elevando ambos os membros da equação ao cubo, teremos que: $(sen^2x + cos^2x)^3 = (1)^3$ $sen^6x + cos^6x + 3sen^4x.cos^2x + 3sen^2x.cos^4x = 1$ $\frac{7}{12} + 3sen^2x.cos^2x(sen^2x + cos^2x) =1$ $3sen^2x.(1-sen^2x) = 1 - \frac{7}{12} \therefore 3sen^2x - 3sen^4x = \frac{5}{12}$ $36sen^4x -36sen^2x +5 = 0$ $sen^2x = \frac{1}{6} \therefore sen(x) = \pm \frac{\sqrt 6}{6}$ ou $sen^2x = \frac{5}{6} \therefore sen(x) = \pm \frac{\sqrt 30}{6}$ $x = arcsen(\pm \frac{\sqrt 6}{6}) + 2k\pi$ ou $x = arcsen(\pm \frac{\sqrt 30}{6}) + 2k\pi$

$$\sin^6 x + \cos^6 x~=~(\sin^2 x)^3 + (\cos^2 x)^3$$$$\sin^6 x + \cos^6 x~=~\underbrace{(\sin^2 x + \cos^2 x)}_{1}\cdot[\underbrace{(\sin^2 x + \cos^2 x)^2}_{1} - 3\sin^2 x \cdot \cos^2 x]$$$$\sin^6 x + \cos^6 x~=~1-\dfrac{3\sin^2 2x}{4}~=~\dfrac{7}{12}$$$$\sin^2 2x~=~\dfrac{5}{9} \implies \cos^2 2x ~=~\dfrac{4}{9} \implies \cos 2x ~=~\pm ~\dfrac{2}{3}$$Pela fórmula do seno do arco-metade, temos: $$\sin x ~=~\pm~\sqrt{\dfrac{1-\frac{2}{3}}{2}~}~=~\pm~\dfrac{1}{\sqrt{6~}} \implies \boxed{x ~=~\pm \arcsin \left(\dfrac{1}{\sqrt{6}}\right)~+~2k\pi}$$$$\sin x ~=~\pm~\sqrt{\dfrac{1+\frac{2}{3}}{2}~}~=~\pm~\dfrac{\sqrt{30~}}{6} \implies \boxed{x ~=~\pm \arcsin \left(\dfrac{\sqrt{30}}{6}\right)~+~2k\pi}$$