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Background - this was part of a homework packet for students looking to skip HS pre-calc. There is a text book they use as well, but this particular problem was not in it.

$$2\sin\left(2x\right)=3\left(1-\cos\left(x\right)\right)$$

My first step was to eliminate the double angle.

$$4\sin\left(x\right)\cos\left(x\right)=3\left(1-\cos\left(x\right)\right)$$

and distribute on right side

$$4\sin\left(x\right)\cos\left(x\right)=3-3\cos\left(x\right)$$

And this is where I am stuck.

Now, I can see that $0$ and $2\pi$ are solutions, but with a graph of both sides, one more. The worksheet instructions do not say whether or not graphing is allowed, although the particular chapter in the book for this seems to rely heavily on calculator work.

My question

Can this be solved by manipulation, if so, how? If not, what is the hint to stop and go to the graph?

enter image description here

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    $\begingroup$ Letting $t=\tan\frac x2$ leads to a polynomial equation with respect to $t$. $\endgroup$ – Alexey Burdin Jul 14 '20 at 15:12
  • $\begingroup$ U need to know will there be an extra root or what the root is $\endgroup$ – Namburu Karthik Jul 14 '20 at 16:50
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    $\begingroup$ @AlexeyBurdin What's the motivation for making the substitution $t = \tan(x/2)$? I've seen Weierstrass substitution in integration but never when solving an equation $\endgroup$ – Ekesh Kumar Jul 14 '20 at 18:58
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Rewrite $2\sin\left(2x\right)=3\left(1-\cos x\right)$ as

$$ 4\sin\frac x2\cos\frac x2\cos x= 3\sin^2\frac x2$$

Then, let $t= \tan\frac x2, \> \cos x = \frac{1-t^2}{1+t^2}$ and factorize

$$\sin\frac x2 \frac{4-3t-4t^2-3t^3}{1+t^2}=0 $$

The factor $\sin\frac x2=0 $ yields $\frac x2 =\pi n$ and $4-3t-4t^2-3t^3=0$ has one real root at

$$t= \frac19\left(-4+\sqrt[3]{584+9\sqrt{4227}}-\sqrt[3]{-584+9\sqrt{4227}}\right) $$

Thus, the solutions are

$$x=2\pi n,\>\>\> 2\pi n + 2\arctan t$$

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  • $\begingroup$ I really appreciate this. I’ve never run into this level of complexity for this particular class level, so while I tip my hat to you, my instruction to the student will be that I expect the intent was to solve via graphing, as I did or on a calculator. $\endgroup$ – JTP - Apologise to Monica Jul 15 '20 at 2:45

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