solve ( 3 sin ( 2 t ) - 1 cos ( t ) = 0 ) for all solutions ( 0 leq t < 2 pi )\n( t = )\ngive your answers…

Explanation:

Step1: Use double - angle formula

Recall the double - angle formula (\sin(2t)=2\sin(t)\cos(t)). The equation (3\sin(2t)-\cos(t) = 0) becomes (3\times(2\sin(t)\cos(t))-\cos(t)=0). [6\sin(t)\cos(t)-\cos(t)=0] Factor out (\cos(t)): (\cos(t)(6\sin(t)-1)=0)

Step2: Solve (\cos(t)=0)

If (\cos(t)=0), then (t=\frac{\pi}{2}+k\pi), (k\in\mathbb{Z}). For (0\leq t\lt2\pi), when (k = 0), (t=\frac{\pi}{2}\approx1.57); when (k = 1), (t=\frac{3\pi}{2}\approx4.71)

Step3: Solve (6\sin(t)-1 = 0)

If (6\sin(t)-1=0), then (\sin(t)=\frac{1}{6}). Using the inverse - sine function (t=\sin^{-1}(\frac{1}{6})+2k\pi) or (t=\pi-\sin^{-1}(\frac{1}{6})+2k\pi), (k\in\mathbb{Z}). (\sin^{-1}(\frac{1}{6})\approx0.17), (\pi-\sin^{-1}(\frac{1}{6})\approx3.14 - 0.17=2.97)

Answer:

(0.17,1.57,2.97,4.71)