Solving Equations: Exact and Decimal Forms

[Jacobpm64]

Find a solution to the equation if possible.
Give the answer in exact form and in decimal form.

1 = 8cos(2x + 1) - 3
1 = 8tan(2x + 1) - 3

I don't know how to do this one.. but I know how to do the simpler ones like..

2 = 5sin(3x)
2/5 = sin(3x)
(sin^-1(2/5)) / 3 = 0.137172

Those are easier... but the one I've said before...

1 = 8cos(2x + 1) - 3 is tough..
1/8 = cos(2x + 1) - 3
3.125 = cos(2x + 1)
{[cos^-1(3.125)] / 2} - 1

Is this correct thus far, and if it is... cos^-1(3.125) is nonreal. So I guess there are no solutions...

How about this one?

1 = 8tan(2x + 1) - 3
1/8 = tan(2x + 1) - 3
3.125 = tan(2x + 1)
{[tan^-1(3.125)] / 2} - 1 = -0.369453

But, the right side doesn't evaluate to 1 when you plug -0.369453 in for x

I don't know where to go with these.

 

[xman]

First, this is a "calc and beyond" and this is a trig question. Second, call the argument something like u. The problem "looks" a whole lot easier this way. Next try chaning the tangent function into another form, namely sines and cosines. You'll find the answer falls out rather quickly.

 

[TD]

1 = 8cos(2x + 1) - 3 is tough..
1/8 = cos(2x + 1) - 3

Ouch! You probably learned that multiplying by 8 on one side, is dividing by it on the other side.
However: there's the -3 that has to go first! It's better to remember the rules:
- you may multiply both sides with a non-zero number
- you may add the same number to both sides

Now, if you'd want to get rid of the 8, you'd have to multiply both sides with 1/8, but that would also give -3/8 on the right side, and no longer a -3.
What you would probably do is first add 3 to both sides (i.e. changing sides of -3), giving 1+3 = 8cos(2x + 1). Now you can remove the 8, giving:

4/8 = cos(2x + 1) <=> cos(2x + 1) = 1/2.

And a cosine can surely be 1/2, so this will have a solution, actually two.
You're making the same mistake with the tan-equation.

 

[Jacobpm64]

haha, sometimes it's fun to laugh at our own stupidity...

Why did I think the right side was a monomial for some reason? I guess I just didn't look at it like I should have. This is a lot easier than I was making it out to be.

Thanks for pointing out my idiocy though lol.

 

[Jacobpm64]

so, let's try that again...

1 = 8cos(2x + 1) - 3
4 = 8cos(2x + 1)
1/2 = cos(2x + 1)

Where do I go from here? .. I'm not sure

{[cos^-1(1/2)] / 2} - 1 = -0.476401

But that answer doesn't work when plugged back in for x..

 

[Tomsk]

You made the same mistake again, you got to cos^-1(1/2)=2x+1 OK, but then you divided by 2 without dividing the 1!

 

[HallsofIvy]

cos(u)= 1/2 gives $u= \frac{2\pi}{3}$ as principal value.
Now, you have $2x+1= \frac{2\pi}{3}$. Solve that for x.