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Upon solving this equation:

$$ 5\sin x - 5\cos x = 2 $$

for the interval $0 \le x < 360^\circ$, I manipulated it in order to get a solvable trigonometric equation below:

$$ \sin 2x = 21/25 $$

Normally this would procure 4 solutions: $28.6^\circ, 61.4^\circ, 208.6^\circ, 241.4^\circ$ (one can easily plot this into Desmos to see that it is true).

However, the answers only include the middle 2 solutions: $61.43^\circ, 208.57^\circ$. Why is this? Why do 2 of the solutions simply disappear? Could it have something to do with the way in which I manipulated the original equation (by squaring and substituting using the appropriate trigonometric identities)?

After looking at the original equation, it makes sense to some degree, but then I'm not a fan of such methods, especially since one could encounter more complicated equations in the future.

Am I missing/forgetting something fundamental?

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    $\begingroup$ How did you manipulate the original equation to get your $4$ answers ? $\endgroup$
    – Hosam H
    Jun 2 at 15:06
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    $\begingroup$ You do not need to use a double angle here. Recall that $\cos x=\sin (\frac{\pi}{2}-x)$ and apply difference to product formula. $\endgroup$
    – Vasili
    Jun 2 at 15:14

1 Answer 1

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$$\sin{x}-\cos{x}=\frac{2}{5}$$ $$(\sin{x}-\cos{x})^2=\frac{4}{25}$$ $$1-2\sin{x}\cos{x}=\frac{4}{25}$$ $$\sin{2x}=\frac{21}{25}$$

In the second step, when we square both sides, we inadvertently include the solutions of: $$\sin{x}-\cos{x}=-\frac{2}{5}$$ This equation is true for $28.6^{\circ}$ and $241.4^{\circ}$, which are the extra two solutions which we don't need.

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  • $\begingroup$ Thanks, this makes total sense. That was the exact mistake I made, by squaring both sides, and forgot that this would invoke the existence of the other 2 solutions. Appreciate the help! $\endgroup$ Jun 15 at 18:37

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