Help with Solving for Variable

**dm10** · October 31st 2008, 05:30 AM

I'm in a hurry and I need some help with solving for the variable in these functions. I've already tried doing them all myself but I can't get the answer.

Here are the problems:

1. -t^(1/3) + (8/3) - 1/3(t) = 0

2. -2sint + 2cos2t = 0

3. e^(x^2/8)-x^(6/8)8*e^(x^2/8)

**Soroban** · October 31st 2008, 07:54 AM

Hello, dm10!

Check the accuracy of your typing.
And it would help if you gave us the answers you're expecting.

$1)\;\;-t^{\frac{1}{3}} + \frac{8}{3} - \frac{1}{3}t\:=\:0$

Multiply by -3: . $3t^{\frac{1}{3}} - 8 + t \:=\:0 \quad\Rightarrow\quad t + 3t^{\frac{1}{3}} - 8 \;=\;0$

Let $u = t^{\frac{1}{3}} \quad\Rightarrow\quad u^3 = t$

Substitute: . $u^3 + 3u - 8 \:=\:0$

But this equation has no rational roots . . .

$2)\;\;-2\sin t + 2\cos2t \:=\: 0$

Identity: . $\cos2\theta \:=\:1-2\sin^2\theta$

So we have: . $2\sin t + 2(1 - 2\sin^2\!t) \:=\:0$

. . which simplifies to: . $2\sin^2\!t + \sin t - 1 \:=\:0$

. . which factors: . $(\sin t + 1)(2\sin t - 1) \:=\:0$

And we have: . $\begin{array}{ccccccccc}\sin t +1\:=\:0 & \Rightarrow & \sin t \:=\:\text{-}1 & \Rightarrow & t \:=\:\frac{3\pi}{2} \\ <br /> 2\sin t-1\:=\:0 & \Rightarrow & \sin t \:=\:\frac{1}{2} & \Rightarrow & t \:=\:\frac{\pi}{6},\:\frac{5\pi}{6} \end{array}$

3. e^(x^2/8)-x^(6/8)8*e^(x^2/8) . . . . What does this say ??

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