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

User SamJL
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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}

Solve for [0,2pi]:15 tan x=5 square root of 3-example-1
User Fractalf
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