Trigonometric equations: the unit circle method

The principle

On $[0, 2\pi]$, each equation has at most 2 solutions. On $\mathbb{R}$, add $+2k\pi$ (period $2\pi$).

$\cos(x) = a$

Solutions: $x = \pm\alpha + 2k\pi$ where $\alpha = \arccos(a)$.

Example: $\cos(x) = \dfrac{1}{2}$ ⇒ $x = \pm\dfrac{\pi}{3} + 2k\pi$.

On the circle: symmetry with respect to the horizontal axis.

$\sin(x) = a$

Solutions: $x = \alpha + 2k\pi$ or $x = \pi - \alpha + 2k\pi$ where $\alpha = \arcsin(a)$.

Example: $\sin(x) = \dfrac{1}{2}$ ⇒ $x = \dfrac{\pi}{6} + 2k\pi$ or $x = \dfrac{5\pi}{6} + 2k\pi$.

On the circle: symmetry with respect to the vertical axis.

$\tan(x) = a$

Solutions: $x = \arctan(a) + k\pi$ (period $\pi$, not $2\pi$).

Example: $\tan(x) = 1$ ⇒ $x = \dfrac{\pi}{4} + k\pi$.

Common special cases

$a$	$\cos(x) = a$	$\sin(x) = a$
$0$	$\pi/2 + k\pi$	$k\pi$
$1/2$	$\pm\pi/3 + 2k\pi$	$\pi/6, 5\pi/6 + 2k\pi$
$\sqrt{2}/2$	$\pm\pi/4 + 2k\pi$	$\pi/4, 3\pi/4 + 2k\pi$
$\sqrt{3}/2$	$\pm\pi/6 + 2k\pi$	$\pi/3, 2\pi/3 + 2k\pi$
$1$	$2k\pi$	$\pi/2 + 2k\pi$

Common mistakes

1. Forgetting $2k\pi$ (must give all solutions on $\mathbb{R}$).
2. Confusing $\cos$ and $\sin$: $\cos$ ⇒ $\pm\alpha$, $\sin$ ⇒ $\alpha$ and $\pi - \alpha$.
3. Missing the second case for $\sin$.
4. Writing $+2k\pi$ for $\tan$ instead of $+k\pi$.

?À toi de jouer

Solutions of sin(x) = √3/2 on [0, 2π]?

A.π/3 and 2π/3B.π/3 onlyC.π/6 and 5π/6D.±π/3

Valider