Question

Practice problem I’m working onThe last one is 2pi/3 and there’s another that you can’t see and it’s pi/4

Answer 1

Given:

`5tan^2\theta+5tan\theta=0`
`-\pi\leq\theta<\pi`

To determine the value(s) for θ based on the given equation, we first let tan θ equal to u. So,

`\begin{gathered} 5tan^(2)\theta+5tan\theta=0 \\ 5u^2+5u=0 \\ Simplify\text{ and rearrange} \\ 5u(u+1)=0 \end{gathered}`

Next, we get the value of u when 5u=0 or u+1=0:

`\begin{gathered} 5u=0 \\ u=0 \end{gathered}`
`\begin{gathered} u+1=0 \\ u=-1 \end{gathered}`

This means that:

`tan\theta=0,\text{ }tan\theta=-1`

Now we consider the given range above. Hence,

For tan θ=0:

`\theta=0,\text{ }\theta=-\pi`

For tan θ=-1:

`\theta=(3\pi)/(4),\theta=-(\pi)/(4)`

Therefore, the answers are:

`\begin{gathered} 0 \\ -\pi \\ -(\pi)/(4) \\ (3\pi)/(4) \end{gathered}`