1. ## [SOLVED] Trigonometry

Question:
Find all the values of $x$ in the interval $0^o \leq x \leq 180^o$ which satisfy the equation $\sin{3x} + 2\cos{3x} = 0.$

Attempt:

$\sin{3x} + 2\cos{3x} = 0.$

$\sin{3x} = -2\cos{3x}$

$\frac{\sin{3x}}{\cos{3x}} = -2$

$\tan{3x} = -2$

$3x = \tan^{-1}(-2)$

Calculator gives no value for $\tan^{-1}(-2)$, need help.

2. Use $\; \tan \left( -x \right) = - \tan \left( x \right)$. Same is true for the inverse.

3. It gives a value for arctan(-2), but it gives a negative result..

Also, I intended to substitute $\sin(3x)$ by $\sqrt{1-\cos^2(3x)}$, but I don't know what it'll give

4. Originally Posted by looi76
Question:
Find all the values of $x$ in the interval $0^o \leq x \leq 180^o$ which satisfy the equation $\sin{3x} + 2\cos{3x} = 0.$

Attempt:

$\sin{3x} + 2\cos{3x} = 0.$

$\sin{3x} = -2\cos{3x}$

$\frac{\sin{3x}}{\cos{3x}} = -2$

$\tan{3x} = -2$

$3x = \tan^{-1}(-2)$

Calculator gives no value for $\tan^{-1}(-2)$, need help.
Originally Posted by xifentoozlerix
Use $\; \tan \left( -x \right) = - \tan \left( x \right)$. Same is true for the inverse.
$\tan{3x} = -2$

$-\tan{3x} = 2$

$-3x = \tan^{-1}(2)$

$x = \frac{\tan^{-1}(2)}{-3}$

$x = 158^o$

Where did I go wrong?

5. $x = \frac{{\tan ^{ - 1} \left( 2 \right)}}{{ - 3}}\approx\frac{{1.10715}}{{ - 3}} \approx - 0.36905$

Now you can see by the original equation that the function has frequency $3$, which means that it's value repeats every $\frac{2\pi}{3}$ radians or every $120^\circ$.

6. Originally Posted by xifentoozlerix
$x = \frac{{\tan ^{ - 1} \left( 2 \right)}}{{ - 3}}\approx\frac{{1.10715}}{{ - 3}} \approx - 0.36905$

Now you can see by the original equation that the function has frequency $3$, which means that it's value repeats every $\frac{2\pi}{3}$ radians or every $120^\circ$.
My mistake was that I didn't configure the calculator to radian mode. It was kept on degree mode. When should I choose the degree mode and when should I choose the radian mode?

7. Originally Posted by looi76
Question:
Find all the values of $x$ in the interval $0^o \leq x \leq 180^o$ which satisfy the equation $\sin{3x} + 2\cos{3x} = 0.$

Attempt:

$\sin{3x} + 2\cos{3x} = 0.$

$\sin{3x} = -2\cos{3x}$

$\frac{\sin{3x}}{\cos{3x}} = -2$

$\tan{3x} = -2$

$3x = \tan^{-1}(-2)$

Calculator gives no value for $\tan^{-1}(-2)$, need help.
Originally Posted by xifentoozlerix
$x = \frac{{\tan ^{ - 1} \left( 2 \right)}}{{ - 3}}\approx\frac{{1.10715}}{{ - 3}} \approx - 0.36905$

Now you can see by the original equation that the function has frequency $3$, which means that it's value repeats every $\frac{2\pi}{3}$ radians or every $120^\circ$.
and can you please tell me all the possible values of $x$ in the interval $0^o \leq x \leq 180^o$

8. Add 120° to -0.369, it'll be between 0 and 180
If you add 120° once again, it'll go over 180. So the only solution is 120-0.369

9. I just realized that the roots repeat in $\frac{2\pi}{6}=\frac{\pi}{3}$. So you will have answers of $x\approx \left[ -0.36905+\frac{\pi}{3},-0.36905+\frac{2\pi}{3},-0.36905+\pi \right]$

10. Thought the period was 2pi, but it's pi...

So add each time pi/3, that is to say 60°

11. But the answer given for the values of x in the textbook is:
$x = 38.9$ , $x = 98.9$ , $x = 158.9$

12. Originally Posted by looi76
But the answer given for the values of x in the textbook is:
$x = 38.9$ , $x = 98.9$ , $x = 158.9$

13. Can I solve this question without using Radians at all?

14. Ok, let's do it again

(in degrees)

$3x=\tan^{-1} (-2) \approx -63.4$

$x \approx -21.1$

Then, add 60 as far as you remain between 0 and 180

It should give you the values you want

15. Originally Posted by looi76
Can I solve this question without using Radians at all?