Question

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

Answer 1

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)`