When dealing with the equation $2\sin2x + \cos2x = 1$, I found that there is two ways of tackling the problem, both yielding different solutions.

The first and most obvious way to tackle the problem is by using the double angle formulae to rewrite the equation, giving the following: $$4\sin x (2\cos x-\sin x) = 0$$ Solving this for solutions in the range $0\le x < 2\pi$ gives $x = 0, \pi, 1.107, 4.249$

The second approach was to rewrite the $\cos2x$ as $\sqrt{1-\sin^22x}$ . This leaves us with:

$2\sin2x + \sqrt{1-\sin^22x} = 1$ , which can be re-arranged to give: $5\sin^2 2x -4 \sin2x = 0$

Factorising and then solving this in the range $0\le x < 2\pi$ gives us the set of solutions $x= 0.464,1.107,3.605,4.249,0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$.

Interestingly, the second method results in double the solutions of the first, and in that set of solutions is the set of solutions attained by use of the first approach. The half of the solutions from the second approach that do not overlap with the set from the first approach are incorrect, meaning that the first approach results in the correct set of solutions.

Why is it that the second approach yields the correct solutions, as well as incorrect solutions?

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    $\begingroup$ I think this is because it's not true that $\cos(2x) = \sqrt{1 - \sin^2(2x)}$, but rather $\pm \cos(2x) = \sqrt{1 - \sin^2(2x)}$. Haven't taken a good look though, just glancing at it however. $\endgroup$ – TrostAft Nov 15 '18 at 22:01
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    $\begingroup$ Squaring doubles possible solutions. $\endgroup$ – herb steinberg Nov 15 '18 at 22:22
  • $\begingroup$ For your range, you have $0° \leq x \lt 360°$, but your answers are in radians. Shouldn't your range be $0 \leq x \lt 2\pi$? $\endgroup$ – bjcolby15 Nov 15 '18 at 23:43
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    $\begingroup$ @bjcolby15 thanks, I've edited that $\endgroup$ – Chris Wilson Nov 15 '18 at 23:58

Consider the equation $x+\sqrt{x}=2$. It can be rearranged as $\sqrt{x}=2-x$, and it becomes $x=4-4x+x^2$ or $x^2-5x+4=0$, so the solutions are $x=1$ or $x=4$.

Oh, wait! If I substitute $x=4$, I get $4+\sqrt{4}=2$, that is, $6=2$. There's something wrong, I guess you agree.

Squaring can introduce spurious roots, in this case also the solution to $$ -\sqrt{x}=2-x $$ that is indeed satisfied for $x=4$.

You're doing the same in your second approach.

Note also that $\cos2x=\sqrt{1-\sin^22x}$ is not true in general.

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    $\begingroup$ Another way to check the answers to an equation like $\sqrt{x} = 2 - x$ is by using the restriction that since $\sqrt{x} \geq 0$, then $2 - x \geq 0$ i.e. $x \leq 2$. $\endgroup$ – 1123581321 Nov 15 '18 at 22:30
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    $\begingroup$ @1123581321 Yes, of course! I just wanted to replicate the OP's way of dealing with the problem in order to show it's wrong. $\endgroup$ – egreg Nov 15 '18 at 22:31

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