General solutions of trigonometric equations: $\sin$, $\cos$ and $\tan$

What this dot point is asking

NESA wants you to write down every solution (over all real numbers) to a trigonometric equation, then restrict to a given interval if asked. A single number like $\frac{\pi}{6}$ is never the whole answer, because $\sin$, $\cos$ and $\tan$ repeat: once you have found one solution, the function's period and its symmetry hand you infinitely many more. The general solution is a compact formula, with an integer parameter $n$, that captures all of them at once. Getting it right means knowing two things for each function: where the second solution in a period sits (its symmetry), and how far apart the repeats are (its period).

The answer

Reading the solutions off the unit circle

Take $\sin x = \frac{1}{2}$. On the unit circle, $\sin x$ is the height of the point, so we draw the horizontal line at height $\frac{1}{2}$ and look for where it meets the circle. The first meeting is at the principal value $\alpha = \arcsin\frac{1}{2} = \frac{\pi}{6}$.

But the line at height $\frac{1}{2}$ cuts the circle a second time, at the point reflected across the vertical axis. That is the angle $\pi - \alpha = \frac{5\pi}{6}$, which has the same height (same sine) but a different angle. This second solution is the one students most often miss.

Each of those two positions repeats every full turn of $2\pi$ (going round the circle again lands on the same point), so we add $2n\pi$ to each. That gives the two infinite families of solutions.

Laid out on a number line, the solutions of $\sin x = \frac{1}{2}$ are two interleaved sequences, each spaced $2\pi$ apart.

General solution for $\sin \theta = k$ (with $-1 \le k \le 1$)

$\sin$ has period $2\pi$ and the reflection symmetry $\sin(\pi - \theta) = \sin\theta$, which is exactly the two-points-per-line picture above. So

where $\alpha = \arcsin k$ is the principal value. A compact single formula for the same set is $\theta = n\pi + (-1)^n \alpha$.

General solution for $\cos \theta = k$ (with $-1 \le k \le 1$)

$\cos$ has period $2\pi$ and is even: $\cos(-\theta) = \cos\theta$. On the unit circle, $\cos x$ is the horizontal coordinate, so a vertical line at $x = k$ meets the circle at an angle and its negative. So

where $\alpha = \arccos k$. The two branches are $+\alpha$ and $-\alpha$, in contrast to sine's $\alpha$ and $\pi - \alpha$. This difference, $\pm\alpha$ for cosine versus $\alpha$ and $\pi - \alpha$ for sine, is the heart of the topic.

General solution for $\tan \theta = k$

$\tan$ has period $\pi$, not $2\pi$, because $\tan(\theta + \pi) = \tan\theta$ (diametrically opposite points on the circle give the same tangent). So there is only one branch:

where $\alpha = \arctan k$. Because $\tan$ takes every real value once per period, $k$ can be any real number (no $-1 \le k \le 1$ restriction).

Equations with a composite argument

If the equation involves $\sin(kx + c)$ rather than $\sin x$, treat the whole bracket as one variable. Solve the general solution for that bracket first, then unwind to $x$ at the very end, dividing every term (including the $2n\pi$) by $k$.

For $\cos 3x = -\frac{\sqrt{3}}{2}$: write $3x = \pm\frac{5\pi}{6} + 2n\pi$, then $x = \pm\frac{5\pi}{18} + \frac{2n\pi}{3}$. The period of $\sin kx$ or $\cos kx$ is $\frac{2\pi}{|k|}$, and of $\tan kx$ is $\frac{\pi}{|k|}$, so a stretched argument produces more closely spaced solutions.

Restricting to a given interval

After writing the general solution, substitute $n = \ldots, -1, 0, 1, 2, \ldots$ and keep only the $\theta$ that fall in the required interval. A number line or quick mental check of each $n$ avoids both missing a solution and including one just outside. Endpoints matter: $[0, 2\pi]$ includes $2\pi$, but $[0, 2\pi)$ does not.

Combining with identities and factoring

Harder equations need a tidy-up before the general solution applies:

• $\sin^2 x = c$: take both square roots, then solve $\sin x = +\sqrt{c}$ and $\sin x = -\sqrt{c}$ separately.
• $a\sin x + b\cos x = c$: convert to $R\sin(x + \alpha) = c$ first (see the auxiliary-angle method), then solve.
• $\sin 2x = \sin x$: bring everything to one side, factor ($2\sin x\cos x - \sin x = \sin x(2\cos x - 1) = 0$), and solve each factor. Never divide by $\sin x$ , that silently deletes the solutions where $\sin x = 0$.

How exam questions ask about general solutions

• "Find the general solution of ...": give the full formula(s) with $n \in \mathbb{Z}$, using both branches for $\sin$/$\cos$.
• "Find all solutions in $[0, 2\pi]$ (or $[0^\circ, 360^\circ]$)": write the general solution, then list the in-range values.
• "Solve $\sin(2x + \ldots) = \ldots$": substitute for the bracket, solve, divide back through.
• "Solve $2\sin^2 x = \ldots$ / a quadratic in $\sin x$": factor or use the quadratic formula in $\sin x$, then apply the general solution to each root.
• "Solve $\sin 2x = \sin x$ / an equation with two trig terms": rearrange and factor; do not cancel a common factor.

Edge cases worth knowing

• Boundary values give a single branch. $\sin x = 1$ has principal value $\frac{\pi}{2}$, but $\pi - \frac{\pi}{2} = \frac{\pi}{2}$ too, so the two sine families merge into the one solution $x = \frac{\pi}{2} + 2n\pi$. Similarly $\cos x = 1$ gives only $x = 2n\pi$.
• No solution when $|k| > 1$ for $\sin$ or $\cos$. $\cos x = 1.4$ has no solutions because cosine never exceeds $1$. Tangent has no such bound.
• A quadratic root that is out of range is discarded. Solving a quadratic in $\sin x$ may give a root like $\sin x = 2$; reject it before writing any general solution.