[SOLVED] Trigonometry

**looi76** · April 22nd 2008, 10:49 AM

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

.

**xifentoozlerix** · April 22nd 2008, 11:00 AM

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

**Moo** · April 22nd 2008, 11:02 AM

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

**looi76** · April 22nd 2008, 11:17 AM

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?

**xifentoozlerix** · April 22nd 2008, 11:37 AM

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

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 $120^\circ$ .

**looi76** · April 22nd 2008, 11:43 AM

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?

**looi76** · April 22nd 2008, 11:46 AM

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 $0^o \leq x \leq 180^o$

**Moo** · April 22nd 2008, 11:48 AM

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

**xifentoozlerix** · April 22nd 2008, 11:51 AM

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

**Moo** · April 22nd 2008, 11:54 AM

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

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

**looi76** · April 22nd 2008, 11:54 AM

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

Can you please explain?

**Isomorphism** · April 22nd 2008, 12:11 PM

Originally Posted by looi76

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

Can you please explain?

Yes that is correct. That is because the others have answered in "radians" and your book has answers in degrees.

**looi76** · April 22nd 2008, 12:13 PM

Can I solve this question without using Radians at all?

**Moo** · April 22nd 2008, 12:16 PM

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

**Isomorphism** · April 22nd 2008, 12:16 PM

Originally Posted by looi76

Can I solve this question without using Radians at all?

Yes. You had previously got the answer as 158.9 degrees.

Now add or subtract 60 degrees to get all other answers.
Adding 60 degrees takes it above 180 degrees, so do not do it.

However

158.9 - 60 = 98.9

98.9 - 60 = 38.9

Thread: [SOLVED] Trigonometry

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