Sabendo que [tex]\sin^2 x + \cos^2 x = 1[/tex], substituímos o valor do seno...
[tex]\sin^2 x + \cos^2 x = 1 \\\\ \frac{1}{16} + \cos^2 x = 1 \\\\ \cos^2 x = \frac{16}{16} - \frac{1}{16} \\\\ \cos^2 x = \frac{15}{16} \\\\ \boxed{\cos x = \pm \frac{\sqrt{15}}{4}}[/tex]
De acordo com o enunciado, x pertence ao 2° quadrante, portanto, negativo: [tex]\boxed{\cos x = - \frac{\sqrt{15}}{4}}[/tex].
Temos que: [tex]\cos(a - b) = \cos a \times \cos b + \sin a \times \sin b[/tex].
Por fim,
[tex]\cos \left ( x - \frac{\pi}{3} \right ) = \\\\ \cos x \times \cos \left ( \frac{\pi}{3} \right ) + \sin x \times \sin \left ( \frac{\pi}{3} \right ) = \\\\ - \frac{\sqrt{15}}{4} \times \frac{1}{2} + \frac{1}{4} \times \frac{\sqrt{3}}{2} = \\\\ \frac{\sqrt{3}}{8} - \frac{\sqrt{15}}{8} = \\\\ \boxed{\boxed{\frac{\sqrt{3} - \sqrt{15}}{8}}}[/tex]
