Помогите

Решите задачу:

$a)\ tg^2\alpha=\frac{sin^2\alpha}{cos^2\alpha} \\\\b)\ 6sin^2\alpha+13cos^2\alpha=10 \\6(1cos^2\alpha)+13cos^2\alpha=10 \\6-6cos^2\alpha+13cos^2\alpha=10 \\7cos^2\alpha=4 \\cos^2\alpha=\frac{4}{7} \\cos\alpha=\pm \sqrt\frac{4}{7} \\\alpha=\pm arccos\sqrt\frac{4}{7}+2\pi n,\n \in Z\ \ \ \ ili\ \ \ \ \alpha=\pm arccos-\sqrt\frac{4}{7}+2\pi n,\n \in Z \\OTBET:\alpha\in\{\pm arccos\sqrt\frac{4}{7}+2\pi n;\ \ \pm arccos-\sqrt\frac{4}{7}+2\pi n\},\ n\in Z$