Solve $$ \sin 2 \theta=-0.4010 \quad 0 \leq \theta \leq 2 \pi $$

Recall the sine function range is between -1 and 1 inclusive, so the value -0.4010 is in range. To find the general solutions for the equation \(\sin 2 \theta = -0.4010\), we first find the inverse sine (also called arcsin) of -0.4010. $$ 2 \theta = \arcsin(-0.4010) $$ The inverse sine function has a range of \([-\pi/2, \pi/2]\), so the first solution \(2 \theta_1\) is in this range. To find the second solution \(2 \theta_2\), we use the symmetry of the sine function about the y-axis: $$ 2 \theta_2 = \pi - \arcsin(-0.4010) $$ Now, we have our two solutions for \(2\theta\).

To find the specific solutions for \(\theta\) in the given range, we first need to divide both the general solutions for \(2\theta_1\) and \(2\theta_2\) by 2: $$ \theta_1 = \frac{1}{2} \arcsin(-0.4010), $$ $$ \theta_2 = \frac{1}{2} (\pi - \arcsin(-0.4010)). $$ Next, we add a multiple of \(\pi\) to \(\theta_2\) until it is in the range \(0 \leq \theta \leq 2\pi\). This is necessary because the sine function has periodicity \(2\pi\), and adding multiples of \(\pi\) will not change the value of \(\theta\) in the equation \(\sin 2\theta = -\,4010\). In this case, we find that the specific solutions for \(\theta\) in the given range are: $$ \theta_1 \approx 3.14159-1.97103=1.17056, $$ $$ \theta_2 \approx 1.97103, $$ Thus, the specific solutions for \(\theta\) in the given range are: \(\theta_1 \approx 1.97103\) and \(\theta_2 \approx 1.17056\).