Question

The equation i must solve is 2* tan(x)^2-18=2
The possible answers are
x = ±arctan(3) + pin, where n is an integer.

x = arctan(3) + 2pin and x = pi + arctan(3) + pin, where n is an integer.

no solutions

x =± arctan(3) + 2pin, where n is an integer.

I believe it to be A because it reduces to +-3 and tangent can add pi. Is this correct.

Answer 1

Whoops, should equal 0 not 2

Answer 2

Is the equation
\[2(\tan x)^2-18=0\]
or
\[2\tan(x^2)-18=0~~?\]

Answer 3

first one

Answer 4

\[\begin{align*}2\tan^2x-18&=0\\
\tan^2x&=9\\
\tan x&=\pm3\\
x&=\arctan(\pm3)\end{align*}\]
Since we're dealing with periodic function with period \(\pi\), tack on an \(n\pi\):
\[x=\arctan(\pm3)+n\pi\]
The inverse tangent function is odd which means \(\arctan (-x)=-\arctan x\), so you can write this as
\[x=\pm\arctan(3)+n\pi\]

Answer 5

so a

Answer 6

Right.