Given the equation:
[tex]\sin (3x)=\sqrt[]{3}\cos (3x)[/tex]Let's find all possible solutions over the interval:
[tex]\lbrack0,2\pi)[/tex]Let first simplify the equation.
Divide both sides by cos(3x):
[tex]\begin{gathered} \frac{\sin3x}{\cos3x}=\frac{\sqrt[]{3}\cos (3x)}{\cos 3x} \\ \\ \frac{\sin 3x}{\cos 3x}=\sqrt[]{3} \end{gathered}[/tex]Apply the trigonometric identity:
sinx/cosx = tanx
We have:
[tex]\tan 3x=\sqrt[]{3}[/tex]Take the inverse tangent of both sides:
[tex]\begin{gathered} 3x=\tan ^{-1}(\sqrt[]{3}) \\ \\ 3x=\frac{\pi}{3} \end{gathered}[/tex]Divide both sides by 3:
[tex]\begin{gathered} \frac{3x}{3}=\frac{\pi}{3}\times\frac{1}{3} \\ \\ x=\frac{\pi}{9} \end{gathered}[/tex]The tangent function is positive in the first and third quadrant. Let's find the second solution by adding π to the reference angle:
[tex]\begin{gathered} 3x=\frac{\pi}{3}+\pi \\ \\ 3x=\frac{\pi+3\pi}{3} \\ \\ 3x=\frac{4\pi}{3} \\ \\ x=\frac{4\pi}{9} \end{gathered}[/tex]Let's find the period of tan(3x):
[tex]\frac{\pi}{b}=\frac{\pi}{3}[/tex]Since the period is π/3, the solution values will repeat every π/3 in both directions.
We have:
[tex]x=\frac{\pi}{9}+\frac{\pi n}{3},\frac{4\pi}{9}+\frac{\pi n}{3}[/tex]When n = 1:
[tex]x=\frac{\pi}{9}+\frac{\pi}{3}=\frac{4\pi}{9}[/tex]When n = 2:
[tex]\frac{\pi}{9}+\frac{2\pi}{3}=\frac{\pi+6\pi}{9}=\frac{7\pi}{9}[/tex]WHen n = 3:
[tex]\frac{\pi}{9}+\frac{3\pi}{3}=\frac{10\pi}{9}[/tex]When n = 4:
[tex]\frac{\pi}{9}+\frac{4\pi}{3}=\frac{\pi+12\pi}{9}=\frac{13\pi}{9}[/tex]When n = 5:
[tex]\frac{\pi}{9}+\frac{5\pi}{3}=\frac{\pi+15\pi}{9}=\frac{16\pi}{9}[/tex]Therefore, the solutions are:
[tex]x=\frac{\pi}{9},\frac{4\pi}{9},\frac{7\pi}{9},\frac{10\pi}{9},\frac{13\pi}{9},\frac{16\pi}{9}[/tex]ANSWER:
[tex]x=\frac{\pi}{9},\frac{4\pi}{9},\frac{7\pi}{9},\frac{10\pi}{9},\frac{13\pi}{9},\frac{16\pi}{9}[/tex]