Question

Solve for [0,2pi]:15 tan x=5 square root of 3

Answer 1

We need to isolate x, then, by moving 15 to the right hand side, we get

`\tan x=\frac{5\sqrt[]{3}}{15}`

which gives

`\tan x=\frac{\sqrt[]{3}}{3}=\frac{1}{\sqrt[]{3}}`

By applying the inverse of the tangent function, we have

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

Now, we need to find which angle corresponds to the inverse tangent of 1 over square root of 3. From the unit circle:

we can see that

`\begin{gathered} x=(\pi)/(6) \\ \text{and} \\ x=(7\pi)/(6) \end{gathered}`

are equivalent to

`\tan ^(-1)(\frac{1}{\sqrt[]{3}})`

Therefore, the answers are

`\begin{gathered} x=(\pi)/(6) \\ \text{and} \\ x=(7\pi)/(6) \end{gathered}`