0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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

Hence

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

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.