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To solve the inverse trigonometric equation $$\cos^{-1}x+\cos^{-1}2x=-\pi,$$ I use the normal cosine addition \begin{align} \cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\\ 2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=\cos(-\pi)\\ 2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=-1\\ (2x^2+1)^2&=(1-x^2)(1-4x^2)\\ 4x^4+1+4x^2&=1-4x^2-x^2+4x^4\\ 9x^2&=0\\ x&=0. \end{align}

Putting $x=0$ in the equation gives LHS $\ne$ RHS: $$\cos^{-1}0+\cos^{-1}0=-\pi \\\pi=-\pi.$$

Since the equation has no solution, why does solving it give zero as a solution? Is my method wrong or is there something else which gives up one solution (i.e. $x=0$ on solving algebraically)?

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Harsh Sharma
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  • Maybe it matters uder what conditions you "may" jump from the one to the next line ... – Math-fun Jul 05 '16 at 12:15
  • @Math-fun Can you please elaborate? I didn't understand that. – Harsh Sharma Jul 05 '16 at 12:17
  • Yes the formula that you are applying must be valid only for some specific values of x which you have completely ignored in your attempt. Hence when we solve such questions we generally do like you did by ignoring the conditions and then put each solution obtained to see if it actually satisfies it or not. It will satisfy the equation if and only if it satisfies the condition which we have ignored. – Matt Jul 05 '16 at 12:19
  • when you square a root equation you change the roots. – Math-fun Jul 05 '16 at 12:19
  • @RaghavSingal Can you please tell me how can I rectify the mistake? – Harsh Sharma Jul 05 '16 at 12:20
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    This equation has no solution, and the reason is trivial: $\cos^{-1} a \ge 0$ for all $a$: in particular LHS is nonnegative and it cannot be equal to $- \pi$. – Crostul Jul 05 '16 at 12:31
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    @HarshSharma you have done a mistake in the beginning you have written : $cos^{-1}(2x^2 +\sqrt {1-x^2} \sqrt{1-4x^2})=-\pi\$ and then in next step you took cosine on both the sides which is wrong as range of cos inverse is (0,$\pi$) and hence you cannot proceed further and you conclude that the equation is not satisfied by any x. – Matt Jul 05 '16 at 12:45
  • @Crostul Thanks a lot. – Harsh Sharma Jul 05 '16 at 13:42
  • @RaghavSingal Thanks! – Harsh Sharma Jul 05 '16 at 13:42

1 Answers1

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\begin{align} \cos^{-1}x+\cos^{-1}2x=-\pi\tag{*}\\ \cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\tag1\\ 2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=-1\tag2\\ (2x^2+1)^2&=(1-x^2)(1-4x^2)\tag3\\ 4x^4+1+4x^2&=1-4x^2-x^2+4x^4\tag4\\ x&=0\tag5. \end{align}

Note that $$\cos^{-1}x+\cos^{-1}2x=\begin{cases}2\pi-&\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&\text{for }x\in[-0.5,0)\\&\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&\text{for }x\in[0,0.5]\end{cases},$$ so step $(1)$ is actually invalid. Fortunately—replacing step $(1)$—the following argument/working is valid: $$\begin{align} &\cos^{-1}x+\cos^{-1}2x =-\pi\tag{*}\\ &\implies \cos(\cos^{-1}x+\cos^{-1}2x) =\cos(-\pi)\\ &\implies 2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}=-1\tag2\\ &\implies (2x^2+1)^2=(1-x^2)(1-4x^2)\tag3\\ &\implies 4x^4+1+4x^2=1-4x^2-x^2+4x^4\tag4\\ &\implies x=0\tag5. \end{align}$$ Therefore, $$\text{equation } (*)\implies x=0.$$

Now, as pointed out by Crostul, since $\arccos$ is a nonnegative function, the given equation $(*)$ is trivially inconsistent (has no solution). Thus, $x{=}0$ is an extraneous solution (it was created in step $(2)\,).$

Since the equation has no solution, why does solving it give zero as a solution?

When a given equation $(A)$ implies equation $(B),$ $(A)$'s solution set (containing the actual solutions) is a subset of $(B)$'s solution set (containing the candidate solutions).

In particular, none of the candidate solutions being an actual solution is necessarily due to deductive explosion, which is necessarily due to equation $(A)$ being inconsistent.

ryang
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