Question

question 3 solve ( 3 sin ( 2 t ) - 1 cos ( t ) = 0 ) for all solutions ( 0 leq t < 2 pi ) ( t = ) give your answers accurate to at least 2 decimal places, as a list separated by commas add work next question

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 \(3\times(2\sin(t)\cos(t))-\cos(t)=0\).
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:

\(1.57,2.97,4.71,0.17\)