The parametric-formulae method

The first method for solving a linear equation $a\sin x + b\cos x + c = 0$ uses the parametric formulae, which express $\sin x$ and $\cos x$ in terms of $t = \tan(x/2)$. In this way the equation becomes rational in $t$.

In summary — Parametric procedure

1. Set $t = \tan(x/2)$.
2. Substitute $\sin x = \dfrac{2t}{1+t^2}$ and $\cos x = \dfrac{1-t^2}{1+t^2}$.
3. Multiply by $1+t^2$ (always positive) to clear the denominator.
4. Solve the rational equation (usually of second degree) in $t$.
5. For each solution $t$, find $x = 2\arctan t + 2k\pi$.
6. Check separately whether $x=\pi$ (that is, $x/2=\pi/2$, where $\tan$ does not exist) is a solution of the original equation, by substituting directly.

Example

Solve $\sqrt{3}\sin x + \cos x + 1 = 0$ with the parametric method.

Substitution: $\sqrt{3}\cdot\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} + 1 = 0 \iff \frac{2\sqrt{3}\,t + 1 - t^2 + 1 + t^2}{1+t^2} = 0$ $\iff 2\sqrt{3}\,t + 2 = 0 \iff t = -\frac{1}{\sqrt{3}}.$

Back to $x$: $\tan(x/2) = -1/\sqrt{3}$, whence $x/2 = -\pi/6 + k\pi$, that is $x = -\dfrac{\pi}{3} + 2k\pi$.

Checking the case $x=\pi$: $\sqrt{3}\sin\pi + \cos\pi + 1 = 0 + (-1) + 1 = 0$. ✓ It too is a solution.

Final solution: $\ x = -\dfrac{\pi}{3} + 2k\pi \ \vee\ x = \pi + 2k\pi$.

Links

Topics: Trigonometric equations
Concepts: Linear trigonometric equation · Parametric formulae · Substitution
Functions: Cosine · Sine · Tangent
Methods: Trigonometric equations
Skills: Solving equations · Using formulae