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This morning I have done for my students of an high school an exercise with the method of the added angle. I not write all the steps but the principals. The equation is:

$$\bbox[5px,border:3px solid #FF7F50]{-2\sin x+\cos x = 1} \tag 1$$

Surely $x=2\Bbb Z \pi$ it is solution of the $(1)$. In fact it is an identity. Being $$\tan \varphi=-\frac 12 \implies \varphi =-\arctan \frac 12, \quad A=\sqrt{5}$$

Now the $(1)$ becomes

$$\bbox[5px,border:3px solid #AA7F50]{\sqrt 5 \sin \left(x-\arctan \frac 12\right)=1} \tag 2$$

Hence I will have the other two solutions:

$$x=\arctan \frac 12+\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi$$ and $$ \quad x=\pi +\arctan \frac 12-\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi \tag 3$$

The solution of my textbook are $x=k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$.

Please, do can be that the $(3)$ are equivalent to $x=2k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$?

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    $\begingroup$ I would say $$ -2\sin x + \cos x =1 $$ is equivalent to $$\sqrt 5\cdot \cos \left( x + \arctan 2\right) =1 $$ or to $$ \sqrt 5\cdot \sin \left( x - \arctan \frac 12\right) =-1$$ $\endgroup$ – dfnu Mar 2 at 13:21
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    $\begingroup$ It could help to realize that $\arctan\frac12 = \arcsin\frac{\sqrt{5}}{5}$. $\endgroup$ – Blue Mar 2 at 13:22
  • $\begingroup$ @dfnu In the meantime, thank you. I did not understand the reason in RHS because you wrote $-1$ when in RHS I have $1$. Everything else is clear. $\endgroup$ – Sebastiano Mar 2 at 21:44
  • $\begingroup$ @Blue I have not understood now $\arctan\frac12 = \arcsin\frac{\sqrt{5}}{5}. Why? Can you put also your answer, please? I like give upvotes :-). $\endgroup$ – Sebastiano Mar 2 at 21:58
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    $\begingroup$ @Sebastiano: Simply consider the $1$-$2$-$\sqrt{5}$ right triangle. ... Since @ user's answer already covers my comment, there's no reason for me to add another. I can wait to get upvotes from you on a later question. :) $\endgroup$ – Blue Mar 2 at 22:10
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The relation: $$ \arctan\frac12=\arcsin\frac1{\sqrt5}=\frac12\arcsin\frac45.\tag1 $$ will help you to compare the results.

Substituting the above values you will find that your result is "almost correct". The spoiler below shows which error gave rise to the deviation from the correct result.

To justify the second equality in (1) one observes the following facts which hold for $\phi=\arcsin\frac1{\sqrt5}$: $$ 0<\frac1{\sqrt5}<\frac1{\sqrt2}\implies 0<\phi<\frac\pi4\implies 0<2\phi<\frac\pi2$$ and $$ \sin2\phi=2\cos\phi\sin\phi=\frac45. $$

Obviously you are using the method of the added angle in the form:$$A\sin x+B\cos x=C(\cos\alpha\sin x+\sin\alpha\cos x)=C\sin(x+\alpha)\tag2$$with$$C=\sqrt{A^2+B^2};\quad\cos\alpha=\frac AC;\quad\sin\alpha=\frac BC.\tag3$$It follows from (3) that $\tan\alpha=\frac BA$ but one should not conclude from this that $\alpha=\arctan\frac BA$ by the following reason. The range of $\arctan x$ is $\left[-\frac\pi2,\frac\pi2\right]$ whereas the equation (3) determines the angle $\alpha$ uniquely in the range $(-\pi,\pi]$. Particularly in your example with$$\cos\alpha=-\frac2{\sqrt5};\quad \sin\alpha=\frac1{\sqrt5},$$the angle $\alpha$ resides in the II quadrant whereas $\arctan\left(-\frac12\right)$ resides in the IV quadrant. Therefore the correct expression for the angle is:$$\alpha=\arctan\left(-\frac12\right)+\pi=\pi-\arctan\frac12,$$and the equation to solve becomes:$$\sqrt5\sin\left(x+\pi-\arctan\frac12\right)=\sqrt5\sin\left(\arctan\frac12-x\right)=1,\tag4$$which solution is:$$x=2\pi k+\begin{cases}\arctan\frac12-\arcsin\frac1{\sqrt5};\\\arctan\frac12+\arcsin\frac1{\sqrt5}-\pi,\end{cases}\tag5$$i.e. differs from your expression by the position of the term $\pi$. Finally with the help of (1) the equation (5) can be written as:$$x=2\pi k+\begin{cases}0;\\\arcsin\frac45-\pi.\end{cases}\tag6$$ It remains to observe that $\alpha=\arcsin\frac45-\pi$ indeed satisfies the relations $\sin\alpha=-\frac45,\,\cos\alpha=-\frac35$. Another issue is the error in the textbook solution (if you cited it correctly): the second solution is $2\pi k$, not $\pi k$.

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  • $\begingroup$ How do you know that it's equal to $\frac{1}{2}\arcsin 0.8$? $\endgroup$ – A-Level Student Mar 2 at 14:26
  • $\begingroup$ @A-LevelStudent $$\sin\left(2\arcsin\frac1{\sqrt5}\right)=2\frac2{\sqrt5}\frac1{\sqrt5}=\frac45.$$ $\endgroup$ – user Mar 2 at 14:29
  • $\begingroup$ Ok, thank you.. $\endgroup$ – A-Level Student Mar 2 at 14:31
  • $\begingroup$ @user Hi, please, can you explain better your hint? Why $\arctan\frac12=\arcsin\frac1{\sqrt5}=\frac12\arcsin\frac45?$ I have understood your comment from $\sin(2\beta)=2\sin\beta\cos\beta$. $\endgroup$ – Sebastiano Mar 2 at 21:56
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    $\begingroup$ @Sebastiano I have added a spoiler with the analysis of an error in your calculation. Hope it will be useful. $\endgroup$ – user Mar 3 at 10:09
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Alternative method:

$$\cos x=1+2\sin x\implies1-\sin^2x=1+4\sin x+4\sin^2x\implies\sin x(4+5\sin x)=0.$$

So the solutions are

$$\sin x=0,\cos x=1+2\cdot0=1$$ $$x=2k\pi$$ and

$$\sin x=-\frac45,\cos x=1+2\left(-\frac45\right)=-\frac35$$ $$x=\arctan\frac43+(2k+1)\pi.$$

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  • $\begingroup$ Hi 😉, when you square the $\cos x=1+2\sin x$ don't you include other solutions that might not be solutions of the initial one? I vote....always up...and thank you very much. $\endgroup$ – Sebastiano Mar 2 at 21:52
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    $\begingroup$ @Sebastiano: good question, but no. Because as you see I plugged back the $\sin$ values in the initial (unsquared) equation to get the $\cos$, and that makes all solutions fully valid. E.g. $\sin x=0$ is accompanied by $\cos x=1$ and not $\cos x=-1$. $\endgroup$ – Yves Daoust Mar 3 at 8:27
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Putting together the two comments we have that the initial equation is equivalent to $$\sqrt 5 \sin\left(x - \arctan \frac12\right) = -1$$ Leading to the solutions $$ x = \arctan \frac12 - \arcsin \frac{\sqrt 5}5 + 2k \pi = 2k \pi$$ and $$x = \pi + \arcsin \frac{\sqrt 5}5 + \arctan \frac12 + 2k \pi=\underbrace{\pi + 2 \arcsin \frac{\sqrt 5}5}_{\alpha}+ 2 k \pi.\tag{1} \label{1}$$ Now let $\alpha = \pi + 2\arcsin \frac{\sqrt 5}5$. We get $$\sin \alpha = -\sin\left( 2\arcsin \frac{\sqrt 5}5\right) = -2 \sin\left( \arcsin \frac{\sqrt 5}5\right)\cos\left(\arcsin\frac{\sqrt 5}5\right)=-\frac45,$$ and similarly for the cosine of $\alpha$.


Note that $k \pi$ is not a solution for odd $k$.

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  • $\begingroup$ Yes....I have seen now and I am done a mistake. +1. Thank you very much...and also for after...😊 $\endgroup$ – Sebastiano Mar 2 at 21:53
  • $\begingroup$ Why $\sqrt 5 \sin\left(x - \arctan \frac12\right) = -1$ if I have $\sqrt 5 \sin\left(x - \arctan \frac12\right) = 1?$ $\endgroup$ – Sebastiano Mar 2 at 22:00
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    $\begingroup$ @Sebastiano I just factored out $\sqrt 5$ and used the expression of sine of the sum of two angles. $\endgroup$ – dfnu Mar 2 at 22:02
  • $\begingroup$ Yes of course but I have $-2\sin x + \cos x =1$ and not $-2\sin x + \cos x =-1$ although $-2\sin x+\cos x = 1\iff 2\sin x - \cos x =1$. I have not mutiply for $-1$ and nothing should change. $\endgroup$ – Sebastiano Mar 2 at 22:06

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