Find the general solution of the trigonometric equation: $a \tan x + b \sin x = c$, where $a,b,c\in \mathbb{R}$ are any real numbers.

I've tried to use the identity $\tan x=\frac{\sin x}{\cos x}$ and then multiplying the original equation by $\cos x$ in order to get the following form of the equation: $a\sin x + b\sin x\cos x = c \cos x$, Also I've tried to divide the original equation by $\sin x$ (supposing that $\sin x \neq 0$) in order to get the following form of the equation: $\frac{a}{\cos x} + b = \frac{c}{\sin x}$, or put in another form: $\frac{c}{\sin x}-\frac{a}{\cos x} = b$, and then I've tried to use a method similar to the method used in the answers given to this question asked here ” Solving trigonometric equations of the form $a\sin x + b\cos x = c$ ” but wasn't able to progress anywhere to something that gives me the general solution.

Thanks for any hint/help.

  • $\begingroup$ Using also $\sin(x)^2+\cos(x)^2=1$ for every real $x$ should do the job. $\endgroup$ – Peter Dec 20 '17 at 12:27

You can use the same strategy as given in this answer in the thread you linked

Let $z=e^{i\theta}$, then $a\tan(\theta) + b\sin(\theta) = c$ is equivalent to

$$ -ia\frac{z -\tfrac 1z}{z + \tfrac 1z}-\tfrac 12 ib(z-\tfrac 1z) = c $$


$$ a(z-\tfrac 1z) + \tfrac 12b(z^2-\tfrac{1}{z^2}) = ic(z+\tfrac 1z)$$

Which, after multiplying by $z^2$, is a polynomial equation of degree $4$, so in principle algebraically solveable. It won't be pretty but if you really want an algebraic solution you will be able to get it this way.


Small thought, not a solution:

When you get to $$ \frac{c}{\sin x}-\frac{a}{\cos x} = b $$ you can put things over a common denominator: $$ \frac{c\cos x}{\sin x \cos x}-\frac{a \sin x }{\sin x \cos x} = b $$ and multiply through to get $$ c\cos x-a \sin x = b\sin x \cos x = \frac{b}{2} \sin(2x) $$ where the left-hand side can be simplified (using an arctangent of $c/a$ or $a/c$ -- I can never remember which) to the form $$ \pm A \sin (x + d) $$ This now gets you an offset sine equal to a double-frequency sine...which really doesn't seem a lot easier to me, but might conceivably be of some use to you.


Another alternative is to say $u = \sin x, v = \cos x$, and let $w$ be a third variable. Then your problem has the form $$ a \frac{u}{v} + bu = c $$ which can be shuffled into $$ au + buv = cv $$ We "homogenize" this by throwing in factors of $w$ to get $$ auw + buv - cvw = 0 $$ and this defines a homogenous polynomial in three unknowns, i.e., it defines a curve on $\Bbb RP^2$. If you can find a point $(u, v, w)$ on this curve, then $(u', v', 1) = (u/w, v/w, 1)$ is a point satisfying the $uv$ form of your equation, but is only of interest to you if $u'^2 + v'^2 = 1$.

Even so, if there happened to be a nice way to parameterize the solution to that homogeneous equation on $\Bbb RP^2$, this might lead to something useful.

  • $\begingroup$ Yes, indeed, by setting $A=\frac{c}{\sqrt{a^2+c^2}}$ and $B=\frac{a}{\sqrt{a^2+b^2}}$ we get that $-1 \leq A,B \leq 1$ and that $A^2+B^2=1$ and thus there exist $ \beta \in \mathbb{R}$ such that $\cos \beta = A$ and $\sin \beta = B$ and we can write the equation in the form $\sqrt{a^2+c^2}\cos(x+\beta)=\sqrt{a^2+c^2}(\cos\beta \cos x - \sin\beta \sin x)=\sqrt{a^2+c^2}(A\cos x - B\sin x)= c\cos x- a\sin x = \frac{b}{2}\sin (2x)$ and the equation stays difficult to solve. $\endgroup$ – MathNerd Dec 20 '17 at 12:50

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