# A series of problems... (don't be afraid it's only maths questions)

• August 6th 2010, 03:27 AM
MathsWhiz
A series of problems... (don't be afraid it's only maths questions)
Ok

Q1) Solve (x^3) - 7x + 6 = 0

Q2) ∫▒〖sec4xtan4x dx〗
if the above isn't readable, i meant find: ∫sec6xtan6x dx

(Giggle) Thanks for any help xD (Lipssealed)
• August 6th 2010, 03:40 AM
Prove It
$f(x) = x^3 - 7x + 6$.

It's pretty easy to see that $f(1) = 0$, so $x - 1$ is a factor.

Long dividing gives

$f(x) = (x-1)(x^2 + x - 6)$

$= (x - 1)(x + 3)(x - 2)$.

So if $x^3 - 7x + 6 = 0$

$(x - 1)(x + 3)(x - 2) = 0$

$x - 1 = 0$ or $x + 3 = 0$ or $x - 2 = 0$.

So $x = -3$ or $x = 1$ or $x = 2$.
• August 6th 2010, 03:42 AM
HallsofIvy
Quote:

Originally Posted by MathsWhiz
Ok

Q1) Solve (x^3) - 7x + 6 = 0

Have you tried anything at all? For example, have you looked at what happens if x= 1?

Quote:

Q2) ∫▒〖sec4xtan4x dx〗
if the above isn't readable, i meant find: ∫sec6xtan6x dx
How did the "4" in the possibly un readable form become "6"?
sec 6x= 1/cos(6x) and tan(6x)= sin(6x)/cos(6x) so this is sin(6x)/cos^2(6x). Let u= cos 6x.

Quote:

Q3) Solve: log(x) + log(x+1) = 1
log(x(x+1)= 1

Quote:

Q4) It is given cos(θ) = 5/13 and that sin(θ) is negative. Find the exact value of tan(θ/2).
$sin^2(\theta)+ cos^2(\theta)= 1$ so you can find $sin(\theta)$ easily and then $tan(\theta)$. Then use the trig identity [tex]tan(\theta)= \frac{2tan(\theta/2)}{1+ tan^2(\theta/2)}. (I actually used $sin(2\theta)= 2sin(\theta)cos(\theta)$ and $cos(2\theta)= cos^2(\theta)- sin^2(\theta)$ to get a formula for $tan(2\theta)$ and then swapped $\theta$ and $2\theta$.

Quote:

Q5) The graphs of y -4x^2 and y = 6 - 2x intersect at x =1. Find the size of the acute angle between these curves at x = 1 correct to the nearest minute.
Find the tangent lines at x= 1 and find the angle between them. You might want to use the trig identity $tan(\theta+ \phi)= \frac{tan(\theta)+ tan(\phi)}{1+ tan(\theta)tan(\phi)}$. Again, I used the sum formulas for sine and cosine to get that formula.

Quote:

i) Express 3cosx - √3sinx in the form R cos (x +a)
$cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(theta)sin(\phi)$. With $\phi= x$, what must $\theta$ be?

Quote:

(Giggle) Thanks for any help xD (Lipssealed)
And the "giggle" and "lipssealed" emoticons are to indicate that anyone giving you answers will be helping you cheat?

In future, try yourself and show what you have tried.
• August 6th 2010, 03:43 AM
Prove It
$\log(x) + \log(x + 1) = 1$

$\log[x(x + 1)] = 1$

$\log(x^2 + x) = 1$

$x^2 + x = e$

$x^2 + x - e = 0$

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-e)}}{2(1)}$

$= \frac{-1 \pm \sqrt{1 + 4e}}{2}$

So $x = \frac{-1 - \sqrt{1 + 4e}}{2}$ or $x = \frac{-1 + \sqrt{1 + 4e}}{2}$.
• August 6th 2010, 04:02 AM
HallsofIvy
Captain Black, did you edit to remove problems? Why?
• August 6th 2010, 04:27 AM
MathsWhiz
Quote:

Originally Posted by HallsofIvy
Have you tried anything at all? For example, have you looked at what happens if x= 1?

How did the "4" in the possibly un readable form become "6"?
sec 6x= 1/cos(6x) and tan(6x)= sin(6x)/cos(6x) so this is sin(6x)/cos^2(6x). Let u= cos 6x.

log(x(x+1)= 1

$sin^2(\theta)+ cos^2(\theta)= 1$ so you can find $sin(\theta)$ easily and then $tan(\theta)$. Then use the trig identity [tex]tan(\theta)= \frac{2tan(\theta/2)}{1+ tan^2(\theta/2)}. (I actually used $sin(2\theta)= 2sin(\theta)cos(\theta)$ and $cos(2\theta)= cos^2(\theta)- sin^2(\theta)$ to get a formula for $tan(2\theta)$ and then swapped $\theta$ and $2\theta$.

Find the tangent lines at x= 1 and find the angle between them. You might want to use the trig identity $tan(\theta+ \phi)= \frac{tan(\theta)+ tan(\phi)}{1+ tan(\theta)tan(\phi)}$. Again, I used the sum formulas for sine and cosine to get that formula.

$cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(theta)sin(\phi)$. With $\phi= x$, what must $\theta$ be?

And the "giggle" and "lipssealed" emoticons are to indicate that anyone giving you answers will be helping you cheat?

In future, try yourself and show what you have tried.

If I knew how to do them, if I knew where to start, I wouldn't be asking for help would I?
• August 6th 2010, 05:05 AM
Prove It
Quote:

Originally Posted by MathsWhiz
If I knew how to do them, if I knew where to start, I wouldn't be asking for help would I?

The point is that you shouldn't be asking for people to do all the work for you without even attempting to do these problems yourself. It can be considered cheating.
• August 6th 2010, 05:42 AM
mr fantastic