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$\sec(x/2) = \cos(x/2)$

I worked on this and got here...

(Let (x/2) = u)

$\cos u - \sec u = 0$

$\cos u(1 - \sec^2u) = 0$

$\cos u[ -1(-1 + \sec^2u)] = 0$

$\cos u(-\tan^2u) = 0$

So, the solutions would be:

$x = pi + 4\pi k, 3\pi + 4\pi k, 0 + 2\pi k$ but the problem is that the first two $(\pi + 4\pi k,3\pi + 4 \pi k)$end up making the original equation have an undefined term $( \sec(x/2))$. Is this simply because I went out of terms of the original equation? If so does this mean that every time I go out of terms of the original I must check the answers? This is confusing me a lot because usually you don't have to check answers unless you square both sides.

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  • $\begingroup$ Introducing "squaring" into the solution of an equation is always dangerous because it "wipes out" the signs of numbers. You can make this a bit easier by considering that you are working with cosine and its reciprocal function, secant. What real numbers are the same as their reciprocals? Can those be values of a cosine and a secant function? If so, what angles are they the cosine for? Those angles are $ \ \frac{x}{2} \ $ , so what values does $ \ x \ $ have? $\endgroup$ Apr 9, 2014 at 1:24
  • $\begingroup$ @RecklessReckoner: Eh? The OP didn't introduce squaring. $\endgroup$
    – MPW
    Apr 9, 2014 at 1:34
  • $\begingroup$ OK, I see what OP did now. You are correct in your answer about the source of the trouble. The derivation could be ended at $ \ \cos u \ (1 - \sec^2 u) = 0 \ , \ \cos u \neq 0 \ \Rightarrow \ \sec^2 u = 1 \ \Rightarrow \ \cos^2 u = 1 \ $ . $\endgroup$ Apr 9, 2014 at 1:43

4 Answers 4

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Hint: $\displaystyle \sec x = \frac{1}{\cos x}$. This means that $$\sec \frac{x}{2} = \cos \frac{x}{2}$$ is equivalent to saying $$\frac{1}{\cos \frac{x}{2}} = \cos \frac{x}{2}$$ or $$\cos^2 \frac{x}{2} = 1$$

Can you take it from there?

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You solved the problem correctly in your original post.

You found all possible potential solutions, and you recognized that some of them are not really solutions. The true solutions are the values $$2\pi k$$ for integral $k$. The false solutions were introduced by setting $\cos u=0$, but that is really implicitly forbidden by the presence of $\sec u$ in the original equation.

Good for you.

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$$\frac{1}{\cos{\frac{x}{2}}}=\cos{\frac{x}{2}}$$ $$\cos^2{\frac{x}{2}}=1$$ $$\sin^2{\frac{x}{2}}=0$$ $$|\sin{\frac{x}{2}}|=0$$ $$\sin{\frac{x}{2}}=0$$ $$\frac{x}{2}=k\pi$$ $$x=2k\pi$$

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As $\displaystyle\cos^2\frac x2=1,\cos x=2\cos^2\frac x2-1=1=\cos0$

$\displaystyle\implies x=2n\pi\pm0$ where $n$ is any integer


Another way : $\displaystyle\cos^2\frac x2=\cos^2\frac A2$

$\displaystyle\implies \frac x2=m\pi\pm\frac A2\iff x=2m\pi\pm A$ where $m$ is any integer

Here $A=0$

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