# Too Many Solutions: tan x = 3 sin x

• Oct 12th 2011, 02:23 AM
HumanShale
Too Many Solutions: tan x = 3 sin x
I approached it this way:

1.) tan x = 3sin x
2.) sin x / cos x = 3sin x
3.) sin x = 3sin x cos x
4.) sin x = 3sin x (sqrt(1 - sin^2 x))
5.) 1 = 3(sqrt(1 - sin^2 x))
6.) 1 = 9(1 - sin^2 x)
7.) 9 -9sin^2 x = 1
8.) -9sin^2 x = -8
9.) sin^2 x = 8/9
10.) sin x = sqrt(8/9)

This gives me solutions in radians at approx: 1.23, 1.91, 5.05, 4.37

The original equation (ignoring solutions at 0, pi, 2pi) has only 2 solutions at: 1.23 and 5.05.

I've drawn graphs at various stages and I start getting 4 solutions at step 4.

What am I doing wrong here?

Thanks.
• Oct 12th 2011, 02:36 AM
mr fantastic
Re: Too Many Solutions: tan x = 3 sin x
Quote:

Originally Posted by HumanShale
I approached it this way:

1.) tan x = 3sin x
2.) sin x / cos x = 3sin x
3.) sin x = 3sin x cos x
4.) sin x = 3sin x (sqrt(1 - sin^2 x))
5.) 1 = 3(sqrt(1 - sin^2 x))
6.) 1 = 9(1 - sin^2 x)
7.) 9 -9sin^2 x = 1
8.) -9sin^2 x = -8
9.) sin^2 x = 8/9
10.) sin x = sqrt(8/9)

This gives me solutions in radians at approx: 1.23, 1.91, 5.05, 4.37

The original equation (ignoring solutions at 0, pi, 2pi) has only 2 solutions at: 1.23 and 5.05.

I've drawn graphs at various stages and I start getting 4 solutions at step 4.

What am I doing wrong here?

Thanks.

I'm not going to wade throgh muddy water when there's a perfectly good alternative route:

sin x = 3sin x cos x

=> sin(x) - 3 sin(x) cos(x) = 0

=> sin(x) (1 - 3 cos(x)) = 0

=> sin(x) = 0 or 1 - 3 cos(x) = 0

etc.
• Oct 12th 2011, 04:55 AM
HumanShale
Re: Too Many Solutions: tan x = 3 sin x
Thank you. Doing so many trig identity questions has got me subbing them in even when they're not needed.

If you can spare the time could you explain to me why my original method started giving me the wrong answers?

Thanks again.
• Oct 12th 2011, 06:25 AM
e^(i*pi)
Re: Too Many Solutions: tan x = 3 sin x
Quote:

Originally Posted by HumanShale
Thank you. Doing so many trig identity questions has got me subbing them in even when they're not needed.

If you can spare the time could you explain to me why my original method started giving me the wrong answers?

Thanks again.

Step 4 -> 5
You cancelled a sin(x) - you can only do this if $\displaystyle \sin(x) \neq 0$. However, 0 is a solution so you've divided by 0

Step 5 -> 6
Squaring introduces extraneous solutions. Be sure to check these in the original equation, better yet avoid it completely.

Step 9 -> Step 10
When taking a square root you need to take the positive and negative roots : $\displaystyle y^2 =x \Leftrightarrow y = \pm x$
• Oct 12th 2011, 06:47 AM
Soroban
Re: Too Many Solutions: tan x = 3 sin x
Hello, HumanShale!

Quote:

$\displaystyle \tan x \,=\,3\sin x$

I approached it this way:

. . $\displaystyle \tan x \,=\,3\sin x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \,=\, 3\sin x \quad\Rightarrow\quad \sin x \,=\, 3\sin x\cos x$

. . $\displaystyle \sin x \,=\,3\sin x\sqrt{1 - \sin^2\!x} \quad\Rightarrow\quad 1 \,=\, 3\sqrt{1 - \sin^2\!x}$
. . . .
You divided by sin x . . . not a good move!

. . $\displaystyle 1 \,=\, 9(1 - \sin^2\!x) \quad\Rightarrow\quad1\,=\, 9 -9\sin^2\!x \quad\Rightarrow\quad 9\sin^2\!x \,=\,8$

. . $\displaystyle \sin^2\!x \,=\,\tfrac{8}{9} \quad\Rightarrow\quad \sin x \,=\, \pm\sqrt{\tfrac{8}{9}} \,=\,\pm\tfrac{2\sqrt{2}}{3}$

This gives me solutions in radians at approx: 1.23, 1.91, 5.05, 4.37

The original equation (ignoring solutions at 0, pi, 2pi)
. . has only 2 solutions at: 1.23 and 5.05.

I've drawn graphs at various stages and I start getting 4 solutions at step 4.

What am I doing wrong here?

Thanks.

By squaring the equation, you risk introducing extraneous roots.

mr fantastic has the best approach.

• Oct 16th 2011, 02:11 AM
HumanShale
Re: Too Many Solutions: tan x = 3 sin x
What a great forum. I've definitely going to have to do some extra work on my trigonometry.