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Practice problem I’m working on The other choices are:Pi/6, pi/3, 2pi/3

Practice problem I’m working on The other choices are:Pi/6, pi/3, 2pi/3-example-1
User Skateboard
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1 Answer

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Given:


\begin{gathered} 4sin^2\theta-4=-3 \\ 0\leq\theta<2\pi \end{gathered}

To determine the value(s) for θ, we follow first let sin θ equal to u. Hence,


\begin{gathered} 4sin^2\theta-4=-3 \\ 4u^2-4=-3 \\ Simplify\text{ and rearrange} \\ 4u^2=-3+4 \\ 4u^2=1 \\ u^2=(1)/(4) \\ u=\pm\sqrt{(1)/(4)} \\ Calculate \\ u=(1)/(2),\text{ u}=-(1)/(2) \end{gathered}

We substitute back u= sin θ. So,


sin\theta=(1)/(2),sin\theta=-(1)/(2)

Now, we consider the range:


\begin{gathered} For\text{ }sin\theta=(1)/(2),0\leq\theta<2\pi: \\ \theta=(\pi)/(6),\theta=(5\pi)/(6) \end{gathered}
\begin{gathered} For\text{ s}\imaginaryI n\theta=-(1)/(2),0\leq\theta\lt2\pi: \\ \theta=(7\pi)/(6),\theta=(11\pi)/(6) \end{gathered}

Therefore, the answers are:


(\pi)/(6),(5\pi)/(6),(7\pi)/(6),(11\pi)/(6)

User Kalinin
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