Solve $$ \cos \left(\frac{\theta-30^{\circ}}{3}\right)=-0.6010 \quad 0 \leq \theta \leq 720^{\circ} $$

First, we need to isolate the angle inside the cosine function. The given equation is: $$\cos \left(\frac{\theta - 30^{\circ}}{3} \right) = -0.6010$$

We need to apply the inverse cosine function (cos^{-1}) to both sides of the equation to find the angle inside the parentheses: $$\frac{\theta - 30^{\circ}}{3} = \arccos(-0.6010)$$ Let's denote the angle on the right-side, which is \(\arccos(-0.6010)\), by \(A\). $$\frac{\theta - 30^{\circ}}{3} = A$$ Now, we need to find the value of \(A\). You can use a calculator to obtain the result (make sure your calculator is using degrees): $$A = \arccos(-0.6010) \approx 127.12^{\circ}$$

As the cosine function is periodic with a period of \(360^{\circ}\), we can add multiples of \(360^{\circ}\) to the angle we found, A, without changing the values of the cosine function. Thus, we have: $$A_1 = 127.12^{\circ} + 360^{\circ} \cdot k$$ where \(k\) is an integer such that the resulting values remain within the allowed range for \(\theta\). Since 0 ≤ θ ≤ 720°, we need to find values of \(k\) that satisfy this condition: $$0 \leq \theta - 30^\circ \leq 720^\circ - 30^\circ = 690^\circ$$ Since the angle, A, lies inside the cosine function, it means that 0 ≤ A ≤ 360°, and so, we need to find integer values of \(k\) such that the following conditions are met: $$0 \leq 127.12^{\circ} + 360^{\circ} \cdot k \leq 360^{\circ}$$ For \(k = 0\), we have \(A_1 = 127.12^{\circ}\). For \(k = 1\), we have \(A_2 = 127.12^{\circ} + 360^{\circ} = 487.12^{\circ}\), which is still within the valid range. For \(k = 2\), we have \(A_3 = 127.12^{\circ} + 720^{\circ} = 847.12^{\circ}\), which is outside the valid range. Therefore, there are two possible values for A: \(A_1 = 127.12^{\circ}\) and \(A_2 = 487.12^{\circ}\).

Now, we can use the values of \(A_1\) and \(A_2\) to solve for θ: For \(A_1 = 127.12^{\circ}\), we have: $$\frac{\theta - 30^{\circ}}{3} = 127.12^{\circ}$$ Multiplying both sides by 3, we get: $$\theta - 30^{\circ} = 381.36^{\circ}$$ Adding 30° to both sides, we find: $$\theta_1 = 411.36^{\circ}$$ For \(A_2 = 487.12^{\circ}\), we have: $$\frac{\theta - 30^{\circ}}{3} = 487.12^{\circ}$$ Multiplying both sides by 3, we get: $$\theta - 30^{\circ} = 1461.36^{\circ}$$ Adding 30° to both sides, we find: $$\theta_2 = 1491.36^{\circ}$$

Now, we verify that our solutions are within the given interval, 0 ≤ θ ≤ 720°: For θ1 = 411.36°, 0° ≤ 411.36° ≤ 720°, so this solution is valid. For θ2 = 1491.36°, 0° ≤ 1491.36° ≤ 720°, which is not true, so this solution is invalid. Thus, the valid solution for the given equation is: $$\theta = 411.36^{\circ}$$