Nora is required to solve the following trigonometric equation.
\(9 \sin ^{2} \theta+12 \sin \theta+4=0, \theta \in\left[0^{\circ},
360^{\circ}\right)\)
Nora did the work shown below. Examine her work carefully. Identify any
errors. Rewrite the solution, making any changes necessary for it to be
correct.
\(9 \sin ^{2} \theta+12 \sin \theta+4=0\)
\((3 \sin \theta+2)^{2}=0\)
$$
3 \sin \theta+2=0
$$
Therefore. \(\sin \theta=-\frac{2}{3}\)
Use a colculator.
\(\sin ^{-1}\left(-\frac{2}{3}\right)=-41.8103149\)
So, the reference ongle is 41.8 , to the neorest tenth of
a degree Sine is negotive in quodrants II ond III. The solution in quadront II
is \(180^{\circ}-41.8^{\circ}=138.2\) The solution in quadrant III is
\(180^{\circ}+41.8=221.8\) Therefore, \(\theta=138.2^{\circ}\) ond \(\theta=221.8\),
to the neorest tenth of a degree.