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)

- Aug 6th 2010, 02:27 AMMathsWhizA 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) - Aug 6th 2010, 02:40 AMProve It
$\displaystyle f(x) = x^3 - 7x + 6$.

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

Long dividing gives

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

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

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

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

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

So $\displaystyle x = -3$ or $\displaystyle x = 1$ or $\displaystyle x = 2$. - Aug 6th 2010, 02:42 AMHallsofIvy
Have you tried

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

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Q2) ∫▒〖sec4xtan4x dx〗

if the above isn't readable, i meant find: ∫sec6xtan6x dx

sec 6x= 1/cos(6x) and tan(6x)= sin(6x)/cos(6x) so this is sin(6x)/cos^2(6x). Let u= cos 6x.

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

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Q4) It is given cos(θ) = 5/13 and that sin(θ) is negative. Find the exact value of tan(θ/2).

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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.

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i) Express 3cosx - √3sinx in the form R cos (x +a)

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(Giggle) Thanks for any help xD (Lipssealed)

In future,**try**yourself and show what you have tried. - Aug 6th 2010, 02:43 AMProve It
$\displaystyle \log(x) + \log(x + 1) = 1$

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

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

$\displaystyle x^2 + x = e$

$\displaystyle x^2 + x - e = 0$

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

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

So $\displaystyle x = \frac{-1 - \sqrt{1 + 4e}}{2}$ or $\displaystyle x = \frac{-1 + \sqrt{1 + 4e}}{2}$. - Aug 6th 2010, 03:02 AMHallsofIvy
Captain Black, did you edit to remove problems? Why?

- Aug 6th 2010, 03:27 AMMathsWhiz
- Aug 6th 2010, 04:05 AMProve It
- Aug 6th 2010, 04:42 AMmr fantastic
Thread closed.

- Aug 6th 2010, 04:47 AMmr fantastic