Solution of a trigonometric linear equation with the method of the added angle

Question

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 $\sin \alpha =-4/5$?

Answer 1

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 $\pi k$.

Answer 2

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

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

Answer 3

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 $\alpha$.

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