Real solutions of a cosx + b sinx = k

Question

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!

Answer 1

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 $$ -\pi+\arctan\frac{4\sqrt{21}-6}{3\sqrt{21}+8}\approx-2.625814 $$

Answer 2

Try in the other way.

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

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

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

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

Answer 3

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 $$x_1=2\tan^{-1}\left(-\frac 12 (3 + \sqrt {21}) \right)\approx -2.62581$$ $$x_2=2\tan^{-1}\left(-\frac 12 (3 - \sqrt {21}) \right)\approx +1.33881$$