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, 2) and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is nosolution, indicate "No Solution."sin(3x) = V3cos(3x)

Answer 1

The given equation is:

$\sin (3x)=\sqrt[]{3}\cos (3x)$

Dividing both sides of the equation by cos(3x):

$\frac{\sin(3x)}{\cos(3x)}=\frac{\sqrt[]{3}\cos(3x)}{\cos(3x)}$

Cancel out the common term in the right hand side of the equation.

$\begin{gathered} \frac{\sin(3x)}{\cos(3x)}=\frac{\sqrt[]{3}\cancel{\textcolor{green}{\cos(3x)}}}{\cancel{\textcolor{green}{\cos(3x)}}} \\ \frac{\sin(3x)}{\cos(3x)}=\sqrt[]{3} \end{gathered}$$\begin{gathered} \text{ Using the trigonometric identity }\frac{\sin(3x)}{\cos(3x)}=\tan (3x),\text{ it follows that} \\ \tan (3x)=\sqrt[]{3} \end{gathered}$

Therefore,

$3x=\tan ^{-1}(\sqrt[]{3})$

Hence,

$\begin{gathered} 3x=\frac{\pi}{3}+\pi n\: \\ \text{ Dividing both sides by }3\text{ we have} \\ x=\frac{\pi}{9}+\frac{\pi n}{3} \end{gathered}$

x = π / 9 + (πn)/3

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