Question

assignment 7.5: solving trigonometric equations
score: 60/100 answered: 7/10
question 8
solve ( 6 sin (2 \beta)-8 sin (\beta)=0 ) for all solutions ( 0 leq \beta<2 pi ).
( \beta= )
give your answers accurate to at least 2 decimal places and in a list separated by commas.
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Explanation:

Step1: Use double - angle formula

Recall the double - angle formula \(\sin(2\beta)=2\sin\beta\cos\beta\).
The equation \(6\sin(2\beta)-8\sin\beta = 0\) becomes \(6\times(2\sin\beta\cos\beta)-8\sin\beta=0\).
Simplify it to \(12\sin\beta\cos\beta - 8\sin\beta=0\).
Factor out \(4\sin\beta\): \(4\sin\beta(3\cos\beta - 2)=0\).

Step2: Solve \(\sin\beta = 0\)

If \(\sin\beta=0\), then \(\beta = k\pi\), \(k\in\mathbb{Z}\).
Since \(0\leq\beta<2\pi\), when \(k = 0\), \(\beta=0\); when \(k = 1\), \(\beta=\pi\).

Step3: Solve \(3\cos\beta-2 = 0\)

If \(3\cos\beta-2 = 0\), then \(\cos\beta=\frac{2}{3}\).
Using \(\beta=\cos^{-1}(x)\), \(\beta=\cos^{-1}(\frac{2}{3})\approx0.84\) or \(\beta = 2\pi-\cos^{-1}(\frac{2}{3})\approx5.44\) (because \(\cos(2\pi - x)=\cos x\) and for \(y = \cos x\), \(x\in[0,2\pi]\), if \(\cos x = a\), \(x=\cos^{-1}(a)\) or \(x = 2\pi-\cos^{-1}(a)\) when \(a\in[- 1,1]\))

Answer:

\(0,\pi,0.84,5.44\)