Solving a trigonometric equation from $[0,2\pi]$

Question

$8\cos(x)=-4\sqrt{2}$

Which is equal to $\cos(x)=-\frac{\sqrt{2}}{2}$

From here I don't know where to take it. Any help would be appreciated, thanks

Answer 1

$cos(x)=cos(3\pi/4)\Longleftrightarrow x=\pm3\pi/4+k2\pi (k\in\mathbb{Z})$.

Since $x\in [0,2\pi]$, we have $x=5\pi/4$ or $x=3\pi/4$

Answer 2

1. Think about what value of $\varphi$ in the first quadrant of the trigonometric circle (i.e. $\varphi\in[0,\pi/2]$) gives $\cos(\varphi)=\sqrt{2}/2$.

2. For what quadrant (of trigonometric circle) $\theta$ have to belong in order to make $\cos(\theta)<0$?

3. What are the angles $x$ related to $\varphi$ (from question 1) in the quadrants from question 2? ("Related" in the sense that $\cos(x)=\pm\cos(\varphi)$)

Answer 3

From $$\cos(x) = \frac{-\sqrt{2}}{2} \text{,} $$

we have, for any integer $k$, \begin{align*} x &= \pm \cos^{-1} \left( \frac{-\sqrt{2}}{2} \right) + 2 \pi k \\ &= \pm \frac{3 \pi}{4} + \frac{8 \pi}{4} k\text{.} \end{align*} The set $\frac{3 \pi}{4} + 2 \pi k$ contains $$ \dots, \frac{-13 \pi}{4}, \frac{-5 \pi}{4}, \frac{3 \pi}{4}, \frac{11 \pi}{4}, \frac{19 \pi}{4}, \dots $$ Of these, only $\frac{3 \pi}{4}$ is in $[0,2\pi] = \left[ 0, \frac{8 \pi}{4} \right]$. The set $\frac{-3 \pi}{4} + 2 \pi k$ contains $$ \dots, \frac{-19 \pi}{4}, \frac{-11 \pi}{4}, \frac{-3 \pi}{4}, \frac{5 \pi}{4}, \frac{13 \pi}{4}, \dots $$ Of these, only $\frac{5 \pi}{4}$ is in $[0,2\pi] = \left[ 0, \frac{8 \pi}{4} \right]$. So the only solutions of the equation in the specified interval are $\frac{3 \pi}{4}$ and $\frac{5 \pi}{4}$.

Answer 4

A figure wouldn't hurt, I guess!