Question

Question 2)

15 Points and will give out brainliest. Thanks

Answer 1

Answer:

$30^{\circ},\ 60^{\circ},\ 120^{\circ},\ 150^{\circ},\ 210^{\circ},\ 240^{\circ},\ 300^{\circ},\ 330^{\circ}$

Step-by-step explanation:

1b) The inequality $(7k-20)(k-4)\le 0$ is true for all

$k\le \dfrac{20}{7}\ \text{or}\ k\ge 4$

2) Consider the equation $\tan^22\theta-3=0$

Rewrite it as

$\tan^22\theta=3$

Take square root:

$\tan 2\theta =\sqrt{3}\ \ \ \text{or}\ \ \ \tan2\theta=-\sqrt{3}\\ \\2\theta=\arctan \sqrt{3}+\pi k, \ k\in Z\ \ \ \text{or}\ \ \ 2\theta =\arctan(-\sqrt{3})+\pi k, \ k\in Z\\ \\2\theta=\dfrac{\pi}{3}+\pi k, \ k\in Z\ \ \ \text{or}\ \ \ 2\theta =-\dfrac{\pi}{3}+\pi k, \ k\in Z\\ \\\theta=\dfrac{\pi}{6}+\dfrac{\pi k}{2}, \ k\in Z\ \ \ \text{or}\ \ \ \theta =-\dfrac{\pi}{6}+\dfrac{\pi k}{2}, \ k\in Z$

Find all values of $\theta$ if $0^{\circ}\le \theta\le 360^{\circ}$

When $k=0:$

$\theta =\dfrac{\pi}{6}=30^{\circ}\ \ \text{[Accept]}\\ \\\theta =-\dfrac{\pi}{6}=-30^{\circ}\ \ \text{[Reject]}$

When $k=1:$

$\theta =\dfrac{\pi}{6}+\dfrac{\pi}{2}=30^{\circ}+90^{\circ}=120^{\circ}\ \ \text{[Accept]}\\ \\\theta =-\dfrac{\pi}{6}+\dfrac{\pi}{2}=-30^{\circ}+90^{\circ}=60^{\circ}\ \ \text{[Accept]}$

When $k=2:$

$\theta =\dfrac{\pi}{6}+\pi=30^{\circ}+180^{\circ}=210^{\circ}\ \ \text{[Accept]}\\ \\\theta =-\dfrac{\pi}{6}+\pi=-30^{\circ}+180^{\circ}=150^{\circ}\ \ \text{[Accept]}$

When $k=3:$

$\theta =\dfrac{\pi}{6}+\dfrac{3\pi}{2}=30^{\circ}+270^{\circ}=300^{\circ}\ \ \text{[Accept]}\\ \\\theta =-\dfrac{\pi}{6}+\dfrac{3\pi}{2}=-30^{\circ}+270^{\circ}=240^{\circ}\ \ \text{[Accept]}$

When $k=4:$

$\theta =\dfrac{\pi}{6}+2\pi=30^{\circ}+360^{\circ}=390^{\circ}\ \ \text{[Reject]}\\ \\\theta =-\dfrac{\pi}{6}+2\pi=-30^{\circ}+360^{\circ}=330^{\circ}\ \ \text{[Accept]}$

When $k\ge 5,$ all angles $\theta >360^{\circ}$