Given:
[tex]5\sin 2\alpha+6\sin \alpha=0,\: 0\le\alpha<\: 2\pi[/tex]
To find: the value of the angle
[tex]\alpha[/tex]
Explanation:
Solving the equation, we get
[tex]\begin{gathered} 5\sin 2\alpha+6\sin \alpha=0 \\ 5\sin 2\alpha=-6\sin \alpha \end{gathered}[/tex]
Using the identity,
[tex]\sin 2\alpha=2\sin \alpha\cos \alpha[/tex]
So, we write
[tex]\begin{gathered} 5(2\sin \alpha\cos \alpha)=-6\sin \alpha \\ 5\cos \alpha=-3 \\ \cos \alpha=-\frac{3}{5} \\ \alpha=\cos ^{-1}(-\frac{3}{5}) \\ \alpha=2.214,4.069 \end{gathered}[/tex]
Final answer: The value of the angle is
[tex]\alpha=2.214,4.0689[/tex]
