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Use inverse functions where needed to find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.) tan2 x + tan x − 30 = 0

Respuesta :

I'm assuming you meant:

[tex]tan^{2}x + tanx - 30 = 0[/tex]
Let [tex]u = tan(x)[/tex]

Thus, we can rewrite it into a quadratic.

[tex]u^{2} + u - 30 = 0[/tex]
[tex](u + 6)(u - 5) = 0[/tex]
[tex]u = -6 or u = 5[/tex]

Hence, [tex]tan(x) = -6[/tex] or [tex]tan(x) = 5[/tex]
[tex]x = tan^{-1}(-6)[/tex] or [tex]x = tan^{-1}(5)[/tex]

Rewriting it in general form, we get:
[tex]x = \pi \cdot n + tan^{-1}(5), n \in Z[/tex]
[tex]x = \pi \cdot n - tan^{-1}(6), n \in Z[/tex]

From there, you can find your solutions from [tex][0, 2\pi)[/tex]

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