Solving a Trigonometric Equation with Domain Restrictions

The student's transformation is incorrect. They wrote (0.6\cos(2t) + 0.8\sin(2t) = \sin(2t+\alpha)) with (\alpha = \arctan(0.6/0.8)). However, the correct identity is (a\cos\theta + b\sin\theta = R\cos(\theta - \phi)) or (R\sin(\theta + \phi)) with specific phase. Let's derive: (0.6\cos(2t) + 0.8\sin(2t) = \sin(2t+\alpha)) would expand to (\sin(2t)\cos\alpha + \cos(2t)\sin\alpha). Equating coefficients: (\cos\alpha = 0.8) and (\sin\alpha = 0.6). Then (\alpha = \arctan(0.6/0.8) = \arctan(0.75) \approx 0.6435) rad. So actually the transformation is correct: (0.6\cos(2t) + 0.8\sin(2t) = \sin(2t+\alpha)) with (\alpha = \arctan(0.75)). Wait, check: (\sin(2t+\alpha) = \sin(2t)\cos\alpha + \cos(2t)\sin\alpha). So coefficient of (\cos(2t)) is (\sin\alpha), which should be 0.6, and coefficient of (\sin(2t)) is (\cos\alpha), which should be 0.8. So (\sin\alpha=0.6), (\cos\alpha=0.8), so (\alpha = \arcsin(0.6) = \arccos(0.8) \approx 0.6435) rad. So the student's (\alpha) is correct. However, the student wrote (\alpha = \arctan(0.6/0.8)), which is also 0.6435 rad, so numerically correct. But the student's equation becomes (\sin(2t+\alpha) = 0.4). That is correct: divide by 5: (0.6\cos + 0.8\sin = 0.4), and the left side equals (\sin(2t+\alpha)), so (\sin(2t+\alpha)=0.4). So the method is correct. The critique: the student's method is actually correct, but they must ensure they use the correct quadrant for (\alpha) and consider all solutions in the domain. The error might be in not finding all solutions or in domain restriction. So the answer: The transformation is correct; (\alpha = \arctan(0.75) \approx 0.6435) rad. The student should then solve (\sin(2t+\alpha)=0.4) for (2t+\alpha) in the appropriate range, then solve for (t) in ([0,\pi]).

We have (3\cos(2t)+4\sin(2t)=5\left(\frac{3}{5}\cos(2t)+\frac{4}{5}\sin(2t)\right)=5\sin(2t+\alpha)) where (\cos\alpha=4/5), (\sin\alpha=3/5), so (\alpha=\arctan(3/4)\approx0.6435) rad. Equation: (5\sin(2t+\alpha)=2), so (\sin(2t+\alpha)=0.4). Let (\theta = 2t+\alpha). Since (t\in[0,\pi]), (2t\in[0,2\pi]), so (\theta\in[\alpha, 2\pi+\alpha]). The sine function equals 0.4 at two angles in each (2\pi) period: (\beta=\arcsin(0.4)\approx0.4115) rad and (\pi-\beta\approx2.7301) rad. We need all (\theta) in ([\alpha, 2\pi+\alpha]) such that (\sin\theta=0.4). The general solutions: (\theta = \beta + 2k\pi) or (\theta = \pi-\beta + 2k\pi). Within the interval: For (\beta+2k\pi): k=0 gives (\beta\approx0.4115), which is less than (\alpha\approx0.6435), so not in interval. k=1 gives (\beta+2\pi\approx6.6947), which is in ([\alpha, 2\pi+\alpha]) since (2\pi+\alpha\approx6.9267). So one solution: (\theta_1=\beta+2\pi). For (\pi-\beta+2k\pi): k=0 gives (\pi-\beta\approx2.7301), which is in interval (since >0.6435). k=1 gives (\pi-\beta+2\pi\approx8.0133), which exceeds (2\pi+\alpha\approx6.9267), so not in interval. Thus two solutions for (\theta): (\theta_1=\beta+2\pi) and (\theta_2=\pi-\beta). Then (t = (\theta-\alpha)/2). So (t_1 = (\beta+2\pi-\alpha)/2 \approx (0.4115+6.2832-0.6435)/2 = 6.0512/2 = 3.0256) s. (t_2 = (\pi-\beta-\alpha)/2 \approx (3.1416-0.4115-0.6435)/2 = 2.0866/2 = 1.0433) s. Both in ([0,\pi]). So two solutions. Exact forms: (t = \frac{\arcsin(0.4)+2\pi-\arctan(3/4)}{2}) and (t = \frac{\pi-\arcsin(0.4)-\arctan(3/4)}{2}). Check that (\alpha=\arctan(3/4)). So two solutions.