Question

15 tan^3 x=5 tan x
Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)

Answer 1

\(\bf 15 tan^3(x)=5tan(x)\implies \cfrac{15 tan^3(x)}{5tan(x)}=1\)

Answer 2

simplify, solve for "tan(x)", then get the inverse tangent to both sides

Answer 3

do u sqrt ?

Answer 4

???

Answer 5

yes

Answer 6

\(\bf 15 tan^3(x)=5tan(x)\implies \cfrac{15 tan^3(x)}{5tan(x)}=1\implies 3tan^2(x)=1\\ \quad \\
tan^2(x)=\cfrac{1}{3}\implies tan(x)=\sqrt{\cfrac{1}{3}}\\ \quad \\
tan^{-1}[tan(x)]=tan^{-1}\left(\sqrt{\cfrac{1}{3}}\right)\implies x=tan^{-1}\left(\sqrt{\cfrac{1}{3}}\right)\)

Answer 7

keep in mind that \(\bf \sqrt{\cfrac{1}{3}}\implies \cfrac{\sqrt{1}}{\sqrt{3}}\implies \cfrac{1}{\sqrt{3}}\\ \quad \\
\textit{simplifying that}\\ \quad \\
\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{3}\)

Answer 8

but i have to keep it on the interval of (0,2pi)?

Answer 9

why is therea sudden -1

Answer 10

I doubt you're reading this, but tan^-1 came from when he got "x" alone the in equation:
\[\tan (x) = \sqrt{1/3}\]\[\ x = \tan^{-1} (\sqrt{1/3)}\]
Getting the inverse tangent gets "x" alone, as does getting the inverse sine, of the equation below, get "x" alone :
\[\sin (x) = k\]\[\ x = \sin^{-1} (k) \]
The interval you shouldn't stress so much about, for it is merely asking for basic, base-set values; no redundancy; [\pi = 3\pi = 5\pi\], so why waste time repeating the same "coordinate".
If this question was non-calculator, just think, "What makes tan(x) = sqrt(3)/3 [you get this value after simplifying sqrt (1/3)]".
Tan(x)= sin(x)/cos(x). Due to this, we may refer back to our unit circle, and guess values that may yield sqrt(3)/3.
With the unit circle, we know the following pairs, its swapped forms, along with some negative variations: (0,1) (1/2, sqrt(3)/2) and (sqrt(2)/2, sqrt(2)/2).
Upon analyzing, we see that sqrt(3)/3 happens when (1/2) is divided by sqrt(3)/2; when sine is 1/2 and cos is sqrt(3)/2.
However, since we want a positive sqrt(3)/3, we know that sine and cosine both have to be either positive or negative. (Quadrants 1 and 3).
From then, now we may finally see that the points that satisfy this are pi/6 and 7pi/6.
Sorry that this is wordy, hope it helped. :)