Помогите 3,4 задание

Решите задачу:

$3.\\\cos^2x=1-\sin^2x=1-\left(\frac23\right)^2=1-\frac49=\frac59\\3\cos^2x+2=3\cdot\frac59+2=\frac53+2=1\frac23+2=3\frac23\\\\\\4.\\\;\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-0,64}=\sqrt{0,36}=\pm0,6\\0\ \textgreater \ \alpha\ \textgreater \ \frac\pi2\Rightarrow\sin x\ \textgreater \ 0\\\sin x=0,6\\2ctg\alpha=3\cdot\frac{\cos\alpha}{\sin\alpha}=3\cdot\frac{0,8}{0,6}=3\cdot\frac43=4$