Solve 6 e^x tan(x) - 3 tan(x) + 8 e^x - 4 = 0

[Math wiz ya rite 09]

This problem scares me. First off, I have not had much work with natural logs, (I think that is what the e's are), I also just started rig this year and my algebra skills aren't the best.

Solve for x, where 0<x<pi: 6e^x * tan(x) - 3tan(x) + 8e^x -4 = 0

Could someone please walk me through this problem.

THANKS!

 

[stapel]

Factor in pairs:

. . . . .3 tan(x) [2e^x - 1] + 4 [2e^x - 1] = 0

Then complete the factorization:

. . . . .[2e^x - 1] [3 tan(x) + 4] = 0

...and solve each factor.

Eliz.

 

[Math wiz ya rite 09]

Factor in pairs:

. . . . .[2e^x - 1] [3 tan(x) + 4] = 0

...and solve each factor.

I set each of the factors equal to zero. For the first factor, I did:

. . .\(\displaystyle \L 2e^x\, -\, 1\, =\, 0\)

. . .\(\displaystyle \L e^x\, =\, \frac{1}{2}\)

Now, by definition of logarithms:

. . .\(\displaystyle \L x\, =\, \ln\left(\frac{1}{2}\right)\, =\, \ln\left(\frac{1}{2}\right)\, =\, \ln(2^{-1})\, =\, -\ln{(2)}\)

For the second part I did:

. . .\(\displaystyle \L 3\tan{(x)}\, +\, 4\, =\, 0\)

. . .\(\displaystyle \L \tan{(x)}\, =\, -\, \frac{3}{4}\)

. . .\(\displaystyle \L \arctan{\left(-\frac{3}{4}\right)}\, =\, 0.6435\)

But now how do I satisfy the domain that I posted above (namely, 0 < x < pi)?

Thank you!

 

[stapel]

...how do I satisfy the domain that I posted above (namely, 0 < x < pi)?

I don't understand your question. Since the solution that you derived for the two factors already is between zero and pi, in what manner does it not "satisfy" the domain requirements?

Please clarify. Thank you.

Eliz.

P.S. You will, of course, want to find any "repeats" of the periodic function which happen to fit into the given interval. Hint: There is one more solution.

 

[Math wiz ya rite 09]

-ln(2) = -0.693 which is not in the domain.

through my graphing calc i got 2.14 as another answer by graphing, but algebraiclly i dont know how to.

 

[stapel]

through my graphing calc i got 2.14 as another answer by graphing, but algebraiclly i dont know how to.

One usually uses trigonometry, not algebra, to find solutions to trigonometric equations.

Eliz.