An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.) 3 tan(3θ) − 1 = 0 (b) Find the solutions in the interval [0, 2π).

Respuesta :

[tex]3\tan3\theta-1=0[/tex]

[tex]3\tan3\theta=1[/tex]

[tex]\tan3\theta=\dfrac13[/tex]

Recall that the tangent function has a period of [tex]\pi[/tex] so that

[tex]3\theta=\tan^{-1}\dfrac13+k\pi[/tex]

for any integer [tex]k[/tex]. Then

[tex]\theta=\dfrac13\tan^{-1}\dfrac13+\dfrac{k\pi}3[/tex]

We get 6 solutions in the interval [0, 2π) for [tex]0\le k\le5[/tex],

[tex]\theta\approx0.107[/tex]

[tex]\theta\approx1.154[/tex]

[tex]\theta\approx2.202[/tex]

[tex]\theta\approx3.249[/tex]

[tex]\theta\approx4.296[/tex]

[tex]\theta\approx5.343[/tex]