Question

Solve the trigonometric equation on the interval of [0,2π) below:


`\large{cos(2 \theta + ( \pi)/(3) ) = ( √(3) )/(2) }`
Show your work, thanks! ​

Answer 1

`\displaystyle \theta = (3\pi)/(4),(11\pi)/(12), (7\pi)/(4), (23\pi)/(12)`

Explanation:

we would like to solve the following trigonometric equation on interval of [0,2π)

`\displaystyle \cos \left(2 \theta + (\pi)/(3) \right ) = ( √(3) )/(2)`

remember that,

`\displaystyle \cos(t) = \cos(2\pi - t)`

so the equation has two solutions

`\begin{cases} \displaystyle \cos \left(2 \theta + (\pi)/(3) \right ) = ( √(3) )/(2) \\ \cos \left(2\pi - (2 \theta + (\pi)/(3) )\right ) = ( √(3) )/(2) \end{cases}`

take inverse trig both sides which yields:

`\begin{cases} \displaystyle 2 \theta + (\pi)/(3) = (\pi)/(6) \\ 2\pi - 2 \theta - (\pi)/(3) = (\pi)/(6) \end{cases}`

simplify the second equation:

`\begin{cases} \displaystyle 2 \theta + (\pi)/(3) = (\pi)/(6) \\ (5\pi)/(3) - 2 \theta = (\pi)/(6) \end{cases}`

add period of 2kπ:

`\begin{cases} \displaystyle 2 \theta + (\pi)/(3) = (\pi)/(6) + 2k\pi \\ (5\pi)/(3) - 2 \theta = (\pi)/(6) + 2k\pi\end{cases}`

By making theta subject of the equation we acquire:

`\begin{cases} \displaystyle \theta = (11\pi)/(12) + k\pi \\ \theta = (3\pi)/(4) - k\pi\end{cases}`

since
`k\in\mathbb{Z}` we get:

`\begin{cases} \displaystyle \theta = (11\pi)/(12) + k\pi \\ \theta = (3\pi)/(4) + k\pi\end{cases}`

as we want to Solve the trigonometric equation on the interval of [0,2π) we get

`\begin{cases} \displaystyle \theta = (11\pi)/(12) \\ \theta = (3\pi)/(4) \end{cases} \text{and} \begin{cases} \displaystyle \theta = (11\pi)/(12) + \pi \\ \theta = (3\pi)/(4) + \pi\end{cases}`

by simplifying we acquire:

`\begin{cases} \displaystyle \theta = (11\pi)/(12) \\ \theta = (3\pi)/(4) \end{cases} \text{and} \begin{cases} \displaystyle \theta = (23\pi)/(12) \\ \theta = (7\pi)/(4) \end{cases}`

and we are done!

hence,

`\displaystyle \theta = (3\pi)/(4),(11\pi)/(12), (7\pi)/(4), (23\pi)/(12)`