How do you solve $\dfrac{2}{3\sqrt{2}}=\cos\left(\dfrac{x}{2}\right)$ for $x$ in the interval $0 \leq x \leq 2\pi$?

This comes from a question that I asked before.

I frequently get stumped when confronting these types of trig functions, so I don't know how to solve this. Can you please provide explanation/steps on how to solve it? Thank you.

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    $\begingroup$ I think you want to use $\cos2\theta=2\cos^{2}\theta-1$. $\endgroup$ – user84413 Jan 7 '14 at 0:23
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    $\begingroup$ are you allowed to use a calculator or do you want to find the exact solution in surds? $\endgroup$ – BlackAdder Jan 7 '14 at 0:28
  • $\begingroup$ @BlackAdder I would like to know how to find the exact solution in surds, thanks $\endgroup$ – Emi Matro Jan 7 '14 at 0:29
  • $\begingroup$ Why do you expect there to be a solution in surds? $\endgroup$ – WillO Jan 7 '14 at 0:36
  • $\begingroup$ @WillO you're right, my apologies. I have the answers as $x=0, 3\pi/4,$ and $2\pi$. I just don't know what work to show to get those answers, for example I don't know how to isolate $x$... $\endgroup$ – Emi Matro Jan 7 '14 at 0:41

Here is what I would do to get rid of surds, even though in this case it does not help a lot.

Square both sides and use the fact $2 \cos^2 x = 1+ \cos 2x$ to get $$ \frac{4}{9\cdot 2} = \cos^2(x/2) = \frac{1 + \cos x}2 $$

Hence $$ \cos (x) = -5/9 $$

Just remember that when you solve for $x$, you write $$ x \approx \pm 2.1589 \text{ or } \pm 123.75 ^\circ $$

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For sure, there is an analytical solution for your equation. It is
$x =2 \arccos\frac{\sqrt2}3$
...which is of absolutely no help to you.

What I would suggest is to first graph your function in order to locate the possible roots (number and locations). When this is done, solve your equation using Newton method. In your case, for $0 < x < 2 \pi$, there is a unique solution which is close to $x = 2$. Let us select this as a starting guess and name it $x_{old}$.

Newton scheme writes
$x_{new} = x_{old} - f(x_{old}) / f'(x_{old})$.
So, in your case, the following iterates are : 2.16376, 2.15983. For sure, you could continue iterating until you reach the desired accuracy. Please notice that $f(2.15983)=-1.19191*10^{-6}$ and that the exact solution (given on my second line) is 2.159827297011170501300826....

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  • $\begingroup$ @Peter Phipps. Thanks for editing for the old man ! Cheers. $\endgroup$ – Claude Leibovici Jan 7 '14 at 18:22

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