Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Nov 21, 2023

Find the solutions to $\tan(x + \pi/3) = \sqrt{3}/3$ over $[0, 2\pi]$ .

STEP 1

Assumptions1. We are solving the equation $\tan \left(x+\frac{\pi}{3}\right)=\frac{\sqrt{3}}{3}$ over the interval $[0, \pi]$ . We are looking for the values of $x$ that satisfy the equation

STEP 2

First, we need to simplify the right side of the equation. The value of $\frac{\sqrt{}}{}$ is equivalent to $\tan(\frac{\pi}{6})$ .
So, we can rewrite the equation as $\tan \left(x+\frac{\pi}{}\right)=\tan(\frac{\pi}{6})$

STEP 3

The general solution of the equation $\tan(\theta1) = \tan(\theta2)$ is $\theta1 = \theta2 + n\pi$ , where $n$ is an integer.
So, we can write the equation as $x+\frac{\pi}{3}=\frac{\pi}{6}+n\pi$

STEP 4

olving for $x$ , we get $x=\frac{\pi}{6}-\frac{\pi}{3}+n\pi$

STEP 5

implify the equation to get $x=-\frac{\pi}{}+n\pi$

STEP 6

Now, we need to find the values of $x$ in the interval $[0,2 \pi]$ .When $n=0$ , we get $x=-\frac{\pi}{6}$ , which is not in the interval.
When $n=1$ , we get $x=\frac{5\pi}{6}$ , which is in the interval.
When $n=2$ , we get $x=\frac{11\pi}{6}$ , which is in the interval.
When $n=3$ , we get $x=\frac{17\pi}{6}$ , which is not in the interval.
So, the solutions to the equation are $x=\frac{5\pi}{6}$ and $x=\frac{11\pi}{6}$ .
The correct answer is $\frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$ .

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