Question

1. 7 points solve the following equation. give all solutions in the interval 0,2π. 2tan²x - 1 = 5

Explanation:

Step1: Isolate $\tan^{2}x$

Add 1 to both sides of the equation $2\tan^{2}x - 1=5$. We get $2\tan^{2}x=5 + 1$, so $2\tan^{2}x=6$. Then divide both sides by 2, $\tan^{2}x = 3$.

Step2: Solve for $\tan x$

Take the square - root of both sides, $\tan x=\pm\sqrt{3}$.

Step3: Find solutions for $\tan x=\sqrt{3}$ in $[0, 2\pi]$

We know that $\tan x=\sqrt{3}$ when $x=\frac{\pi}{3}$ and $x=\frac{\pi}{3}+\pi=\frac{4\pi}{3}$ since the tangent function has a period of $\pi$ and $\tan\frac{\pi}{3}=\sqrt{3}$.

Step4: Find solutions for $\tan x =-\sqrt{3}$ in $[0, 2\pi]$

We know that $\tan x=-\sqrt{3}$ when $x=\frac{2\pi}{3}$ and $x=\frac{2\pi}{3}+\pi=\frac{5\pi}{3}$ since the tangent function has a period of $\pi$ and $\tan\frac{2\pi}{3}=-\sqrt{3}$.

Answer:

$x = \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$