Question

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

Answer 1

Answer:

$\displaystyle \theta = \frac{3\pi}{4},\frac{11\pi}{12}, \frac{7\pi}{4}, \frac{23\pi}{12}$

Step-by-step explanation:

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

$\displaystyle \cos \left(2 \theta + \frac{\pi}{3} \right ) = \frac{ \sqrt{3} }{2}$

remember that,

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

so the equation has two solutions

$\begin{cases} \displaystyle \cos \left(2 \theta + \frac{\pi}{3} \right ) = \frac{ \sqrt{3} }{2} \\ \cos \left(2\pi - (2 \theta + \frac{\pi}{3} )\right ) = \frac{ \sqrt{3} }{2} \end{cases}$

take inverse trig both sides which yields:

$\begin{cases} \displaystyle 2 \theta + \frac{\pi}{3} = \frac{\pi}{6} \\ 2\pi - 2 \theta - \frac{\pi}{3} = \frac{\pi}{6} \end{cases}$

simplify the second equation:

$\begin{cases} \displaystyle 2 \theta + \frac{\pi}{3} = \frac{\pi}{6} \\ \frac{5\pi}{3} - 2 \theta = \frac{\pi}{6} \end{cases}$

add period of 2kπ:

$\begin{cases} \displaystyle 2 \theta + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \\ \frac{5\pi}{3} - 2 \theta = \frac{\pi}{6} + 2k\pi\end{cases}$

By making theta subject of the equation we acquire:

$\begin{cases} \displaystyle \theta = \frac{11\pi}{12} + k\pi \\ \theta = \frac{3\pi}{4} - k\pi\end{cases}$

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

$\begin{cases} \displaystyle \theta = \frac{11\pi}{12} + k\pi \\ \theta = \frac{3\pi}{4} + k\pi\end{cases}$

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

$\begin{cases} \displaystyle \theta = \frac{11\pi}{12} \\ \theta = \frac{3\pi}{4} \end{cases} \text{and} \begin{cases} \displaystyle \theta = \frac{11\pi}{12} + \pi \\ \theta = \frac{3\pi}{4} + \pi\end{cases}$

by simplifying we acquire:

$\begin{cases} \displaystyle \theta = \frac{11\pi}{12} \\ \theta = \frac{3\pi}{4} \end{cases} \text{and} \begin{cases} \displaystyle \theta = \frac{23\pi}{12} \\ \theta = \frac{7\pi}{4} \end{cases}$

and we are done!

hence,

$\displaystyle \theta = \frac{3\pi}{4},\frac{11\pi}{12}, \frac{7\pi}{4}, \frac{23\pi}{12}$