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.

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.

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.

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 : $\displaystyle y^2 =x \Leftrightarrow y = \pm x$

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.

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.