Question 150
Determine the closest two values of $x$ (default and excess) to $2003^{\circ}$ that meet the following trigonometric equation: $$
\frac{1}{\operatorname{sen}^{2} x}-\frac{1}{\cos ^{2} x}-\frac{1}{\operatorname{tg}^{2} x}-\frac{1}{\operatorname{cotg}^{2} x}-\frac{1}{\sec ^{2} x}-\frac{1}{\operatorname{cosec}^{2} x}=-3
$$
\frac{1}{\operatorname{sen}^{2} x}-\frac{1}{\cos ^{2} x}-\frac{1}{\operatorname{tg}^{2} x}-\frac{1}{\operatorname{cotg}^{2} x}-\frac{1}{\sec ^{2} x}-\frac{1}{\operatorname{cosec}^{2} x}=-3
$$