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

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

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

• You should at least check which quadrants of $[0,2\pi]$ give a negative value of cosine. In fact the value shown is the cosine of a "special angle" we learn about in high school geometry. – hardmath Oct 3 '18 at 21:59

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

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

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 $$\frac{5 \pi}{4}$$.

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 $\cos(x)=\pm\cos(\varphi)$)

• Using $x$ and $y$ for angles will cause confusion if you're talking about quadrants. – mr_e_man Sep 22 '18 at 0:59
• @mr_e_man I'd no idea.. thanks for pointing this out – Rodrigo Dias Sep 22 '18 at 1:16
• $x$ and $y$ are usually used for the rectangular coordinates; the first quadrant has $x>0,\;y>0$. – mr_e_man Sep 22 '18 at 1:18
• @mr_e_man Hope the edit helps – Rodrigo Dias Sep 22 '18 at 1:28

A figure wouldn't hurt, I guess!