Question

solve ( 5 sin ( 2 phi ) + 1 sin ( phi ) = 0 ) for all solutions ( 0 leq phi < 2 pi ).

( phi = )

give your answers accurate to at least 2 decimal places and in a list separated by commas.

question help: video

Explanation:

Step1: Use double - angle formula

Recall the double - angle formula \(\sin(2\phi)=2\sin\phi\cos\phi\). The equation \(5\sin(2\phi)+\sin\phi = 0\) becomes \(5\times(2\sin\phi\cos\phi)+\sin\phi=0\), which simplifies to \(10\sin\phi\cos\phi+\sin\phi = 0\).

Step2: Factor out \(\sin\phi\)

Factor out \(\sin\phi\) from the left - hand side of the equation: \(\sin\phi(10\cos\phi + 1)=0\).

Step3: Set each factor equal to zero

• Case 1: \(\sin\phi=0\)

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

• Case 2: \(10\cos\phi+1 = 0\)

Solve \(10\cos\phi+1 = 0\) for \(\cos\phi\). We get \(\cos\phi=-\frac{1}{10}=- 0.1\).
Using the inverse - cosine function \(\phi=\cos^{-1}(-0.1)\) or \(\phi = 2\pi-\cos^{-1}(-0.1)\).
Since \(\cos^{-1}(-0.1)\approx1.67\) (using a calculator), then \(2\pi-\cos^{-1}(-0.1)\approx2\times3.14 - 1.67=4.61\).

Answer:

\(0,\pi,1.67,4.61\)