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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 \(\displaystyle x = 2\).

Have you triedOk

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

How did the "4" in the possibly un readable form become "6"?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.

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

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

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

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

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

In future,

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

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

\(\displaystyle x^2 + x = e\)

\(\displaystyle x^2 + x - e = 0\)

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

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

So \(\displaystyle x = \frac{-1 - \sqrt{1 + 4e}}{2}\) or \(\displaystyle x = \frac{-1 + \sqrt{1 + 4e}}{2}\).

\(\displaystyleHave you triedanythingat 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

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

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

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

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

In future,tryyourself 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?\)

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

The post was edited while you and others were replying. The reason for the edit is given in rule #8.Captain Black, did you edit to remove problems? Why?

Regarding this thread: Rule #11 is also relevant. As is rule #4. Rule #6 may also be relevant (hence the current closure of the thread).

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