How do you solve $\sqrt{tan x} = \sqrt[4]{3}$ $sqrt(tanx)=root4(3)$ in the interval [0,360]?

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How do you solve $\sqrt{tan x} = \sqrt[4]{3}$ $sqrt(tanx)=root4(3)$ in the interval [0,360]?

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Answer 1 · 2016-10-25T18:17:06

Rewrite both sides in radical form.

${(tan x)}^{\frac{1}{2}} = 3^{\frac{1}{4}}$ $(tanx)^(1/2) = 3^(1/4)$

We need to balance exponents.

${({(tan x)}^{\frac{1}{2}})}^{4} = {(3^{\frac{1}{4}})}^{4}$ $((tanx)^(1/2))^4 = (3^(1/4))^4$

${(tan x)}^{2} = 3$ $(tanx)^2 = 3$

${tan}^{2} x = 3$ $tan^2x = 3$

$tan x = \pm \sqrt{3}$ $tanx = +-sqrt(3)$

$x = 60^{\circ}, 120^{\circ}, 240^{\circ}, 300^{\circ}$ $x = 60^@, 120^@, 240^@, 300^@$

However, we can only take the solutions when the $\sqrt{3}$ $sqrt(3)$ is positive, since the solutions located in quadrants where tangent is negative will render the original equation undefined.

So, $x = 60^{\circ} and 240^{\circ}$ $x = 60^@ and 240^@$

Hopefully this helps!

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How do you solve $\sqrt{tan x} = \sqrt[4]{3}$ $sqrt(tanx)=root4(3)$ in the interval [0,360]?

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