1. Trig Equation

How do I solve 6cos(3x) = 3 without using the unit circle?

2. $\displaystyle 6\cos(3x) = 3$

$\displaystyle \cos(3x) = \frac{1}{2}$

$\displaystyle 3x = \frac{\pi}{3}$

$\displaystyle x = \frac{\pi}{9}$

3. Are there any restrictions on x? Usually we solve with the restriction $\displaystyle 0 \le x < 2\pi$. If that's the case, then we'll need to expand on pickslides' work a bit. If we have the restriction $\displaystyle 0 \le x < 2\pi$, then it follows that $\displaystyle 0 \le {\color{red}3}x < {\color{red}6}\pi$.

Originally Posted by pickslides
$\displaystyle 6\cos(3x) = 3$

$\displaystyle \cos(3x) = \frac{1}{2}$

$\displaystyle 3x = \frac{\pi}{3}$
Between 0 and 2 pi we also have 5pi/3 as a possible answer:

$\displaystyle 3x = \frac{5\pi}{3}$

For each answer, we need to find coterminal angles to it that is in $\displaystyle 0 \le 3x < 6\pi$, so we add 2pi and 4pi to each:

$\displaystyle 3x = \frac{\pi}{3}$
$\displaystyle 3x = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$
$\displaystyle 3x = \frac{\pi}{3} + 4\pi = \frac{13\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3} + 4\pi = \frac{17\pi}{3}$

Now take each equation and divide both sides by 3 to solve for x. So we'll get
$\displaystyle x = \frac{\pi}{9},\;\frac{5\pi}{9},\;\frac{7\pi}{9},\; \frac{11\pi}{9},\;\frac{13\pi}{9},\;\frac{17\pi}{9 },\;$

These are all possible values of x that is in $\displaystyle 0 \le x < 2\pi$.

01

4. Yeongil is correct to show 6 solutions, although given the following

Originally Posted by magentarita
without using the unit circle?
I would only give the 1 solution using the triangle made with angles $\displaystyle \frac{\pi}{2},\frac{\pi}{3}$ and $\displaystyle \frac{\pi}{6}$ ignoring the unit circle.

5. more...

Originally Posted by pickslides
Yeongil is correct to show 6 solutions, although given the following

I would only give the 1 solution using the triangle made with angles $\displaystyle \frac{\pi}{2},\frac{\pi}{3}$ and $\displaystyle \frac{\pi}{6}$ ignoring the unit circle.
I need more than one or two answers.

6. Exactly...

Originally Posted by yeongil
Are there any restrictions on x? Usually we solve with the restriction $\displaystyle 0 \le x < 2\pi$. If that's the case, then we'll need to expand on pickslides' work a bit. If we have the restriction $\displaystyle 0 \le x < 2\pi$, then it follows that $\displaystyle 0 \le {\color{red}3}x < {\color{red}6}\pi$.

Between 0 and 2 pi we also have 5pi/3 as a possible answer:

$\displaystyle 3x = \frac{5\pi}{3}$

For each answer, we need to find coterminal angles to it that is in $\displaystyle 0 \le 3x < 6\pi$, so we add 2pi and 4pi to each:

$\displaystyle 3x = \frac{\pi}{3}$
$\displaystyle 3x = \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$
$\displaystyle 3x = \frac{\pi}{3} + 4\pi = \frac{13\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}$
$\displaystyle 3x = \frac{5\pi}{3} + 4\pi = \frac{17\pi}{3}$

Now take each equation and divide both sides by 3 to solve for x. So we'll get
$\displaystyle x = \frac{\pi}{9},\;\frac{5\pi}{9},\;\frac{7\pi}{9},\; \frac{11\pi}{9},\;\frac{13\pi}{9},\;\frac{17\pi}{9 },\;$

These are all possible values of x that is in $\displaystyle 0 \le x < 2\pi$.

01
This is exactly what I needed.

7. Hello: Here are the details for the solution :

The general solution is :

I'have :
but :
Calculate :
Case 1 :
but :
¨:

Case 2 :

We find :
¨
:

Thank you for all

8. wow!

Originally Posted by dhiab
Hello: Here are the details for the solution :

The general solution is :

I'have :
but :
Calculate :
Case 1 :
but :
¨:

Case 2 :

We find :
¨
:

Thank you for all
A very detailed reply!