Question

question #13 determine the angle or angles on the unit circle that have cosθ = 1 θ = π/2 θ = 0π and θ = π and θ = 2π θ = 0π and θ = 2π θ = π/2 and θ = 3π/2 question #14 solve the following trigonometric equation for 0 ≤ x ≤ 2π : 9tan(x) - 3√3 = 0. x = π/3, 4π/3 x = π/6, 7π/6 x = 5π/6, 11π/6 x = 2π/3, 5π/3

Explanation:

Step1: Recall cosine on unit - circle

The cosine of an angle $\theta$ in the unit - circle is the $x$ - coordinate of the point on the unit - circle corresponding to the angle $\theta$. We know that $\cos\theta = 1$ when the $x$ - coordinate of the point on the unit - circle is 1. This occurs at $\theta = 0$ and $\theta = 2\pi$ (since angles in the unit - circle are periodic with period $2\pi$).

Step1: Solve the trigonometric equation

Given $9\tan(x)-3\sqrt{3}=0$. First, isolate $\tan(x)$:
$9\tan(x)=3\sqrt{3}$, so $\tan(x)=\frac{3\sqrt{3}}{9}=\frac{\sqrt{3}}{3}$.

Step2: Find the values of $x$ in the given range

We know that $\tan(x)=\frac{\sqrt{3}}{3}$ when $x = \frac{\pi}{6}+k\pi$, where $k\in\mathbb{Z}$. For $0\leq x\leq2\pi$, when $k = 0$, $x=\frac{\pi}{6}$; when $k = 1$, $x=\frac{\pi}{6}+\pi=\frac{7\pi}{6}$.

Answer:

$\theta = 0\pi$ and $\theta = 2\pi$