Question

Use trigonometric identities and algebraic methods, as necessary, to solve the following trigonometric equation. Please identify all possible solutions by including allanswers in (0, pi) and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is no solutionindicate "No Solution."2v3tan(x) – 2 = 0

Answer 1

ANSWER:

General solution:

$x=\frac{\pi}{6}+n\pi$

Solution over the given interval:

$x=\frac{\pi}{6}$

EXPLANATION:

Let's solve for tan(x),

$\begin{gathered} 2\sqrt[]{3}\tan (x)-2=0\rightarrow2\sqrt[]{3}\tan (x)=2 \\ \rightarrow\sqrt[]{3}\tan (x)=1\rightarrow\tan (x)=\frac{1}{\sqrt[]{3}} \\ \\ \Rightarrow\tan (x)=\frac{\sqrt[]{3}}{3} \end{gathered}$

Now, we'll use the inverse trigonometric function for tangent:

$\begin{gathered} \tan (x)=\frac{\sqrt[]{3}}{3}\rightarrow x=\tan ^{-1}(\frac{\sqrt[]{3}}{3}) \\ \\ \Rightarrow x=\frac{\pi}{6} \end{gathered}$

Now, since tan(x) has a period of pi, the general solution for the equation is:

$x=\frac{\pi}{6}+n\pi,n\in Z$

For the interval [0,pi] we'll have the solution:

$x_{}=\frac{\pi}{6}$