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I'm going through Riley's 'Mathematical methods for physics and engineering' on my own and faced something relatively simple, yet I'm unable to figure this out.

The question is the standard problem, to solve for x in:

$3 \sin x - 4 \cos x = 2$

I solve this first dividing both sides by 4 (to make everything ≤1 to be easier to work with) and continue by procedure, described in the book and here: Solving trigonometric equations of the form $a\sin x + b\cos x = c$

I get something like: $x = \arcsin(\frac{2}{5})-\arctan(-\frac{4}{3}) \approx 1.339$, which is correct, as given in the book answer.

However, the book gives 2 answers. The second is $\approx-2.626$ and I am unable to obtain it.

I understand $\arcsin$ and $\arctan$ functions can give two different values, so tried different variations of it, which are:

  1. $\arcsin$ to give the angle in the 2nd quadrant, by $x_2=\pi-\arcsin(\frac{2}{5})$
  2. $\arctan$ to give the angle in the 2nd quadrant (my calculator gives $\arctan(-\frac{4}{3})\approx -0.927$) by adding a $\pi$

However, neither of these combinations work to give me the required $-2.626$.

So my questions are:

  1. How would you obtain the required second answer?
  2. Why other combinations from $\arcsin$ and $\arctan$ wouldn't work?
  3. How could I solve this (if at all) without dividing primarily by the highest coefficient, in this case 4? It somehow just didn't work in my head.

Many thanks and let me know if you need any further clarification!

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  • $\begingroup$ Welcome to the site ! $\endgroup$ – Claude Leibovici Jul 27 at 13:58
  • $\begingroup$ This answer of mine to the "$a \sin x + b\cos x = c$" question you link gives a diagrammatic solution. It notes a sign ambiguity in the value it calls $d$, but doesn't mention that this leads to two solutions. This other answer of mine illustrates both solutions. Perhaps these will aid your intuition. $\endgroup$ – Blue Jul 27 at 14:11
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That's a problem with “mechanical” solutions. Transform the problem into $$ \begin{cases}3Y-4X=2 \\[4px] X^2+Y^2=1 \end{cases} $$ by setting $X=\cos x$, $Y=\sin x$. Then we can write $3Y=4X+2$ and therefore $$ 9X^2+16X^2+16X+4=9 $$ that becomes $25X^2+16X-5=0$. This leads to $$ X=\dfrac{-8+3\sqrt{21}}{25} \qquad\text{or}\qquad X=\dfrac{-8-3\sqrt{21}}{25} $$ and the solutions are $$ \begin{cases} X=\dfrac{3\sqrt{21}-8}{25} \\[6px] Y=\dfrac{4\sqrt{21}+6}{25} \end{cases} \qquad\text{or}\qquad \begin{cases} X=-\dfrac{3\sqrt{21}+8}{25} \\[6px] Y=-\dfrac{4\sqrt{21}-6}{25} \end{cases} $$ The first solution is in the first quadrant, so we can evaluate it as $$ \arctan\frac{4\sqrt{21}+6}{3\sqrt{21}-8}\approx1.338812 $$ The second solution is in the third quadrant, so we can evaluate it as $$ -\pi+\arctan\frac{4\sqrt{21}-6}{3\sqrt{21}+8}\approx-2.625814 $$

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  • $\begingroup$ Thanks a lot. Very useful. $\endgroup$ – Nigi Tanka Jul 29 at 8:24
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Try in the other way.

$$4\cos x = -2+3\sin x \implies 16\cos^2x = 4+9\sin^2x-12\sin x$$ $$ \implies 16 - 16\sin^2x = 4+9\sin^2x-12\sin x$$

$$25\sin^2x-12\sin x -12 = 0 $$

$$\sin x \approx -0.493,0.973 \text{ both have magnitude }\le1 $$

$$x = \arcsin(0.973) = \color{blue}{1.339} \ , x = \arcsin(-0.492) = -0.514 \color{blue}{\equiv -(0.514 + \pi) = -2.627}$$

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  • $\begingroup$ Thanks a lot. Couldn't accept more than one answer as 'accepted', but this was very useful too. $\endgroup$ – Nigi Tanka Jul 29 at 8:30
  • $\begingroup$ You're welcome. No worries : ) $\endgroup$ – Ak19 Jul 29 at 11:53
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Using the tangent half-angle substitution, you would need to solve $$t^2+3 t-3=0 \implies t=-\frac 12 (3 \pm \sqrt {21})$$ making $$x_1=2\tan^{-1}\left(-\frac 12 (3 + \sqrt {21}) \right)\approx -2.62581$$ $$x_2=2\tan^{-1}\left(-\frac 12 (3 - \sqrt {21}) \right)\approx +1.33881$$

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  • $\begingroup$ Thanks, solved this way too. However, I'd think that having no information about sinx and cosx values within the tanx, it's a limitation, regarding to within which quadrant the real answer is. On the other hand, the solution gives exactly the right answers. Is this a coincidence, or it will always give me the right solution? I couldn't think of the way to prove it. $\endgroup$ – Nigi Tanka Jul 29 at 8:28

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