Solving trig equations — general solutions, solutions in given range: Visual walkthrough

Step 1 — What is an angle, and what is a "sine"?

WHAT. Draw a circle of radius $1$ centred at the origin. This is the unit circle (see Unit Circle and Radian Measure). Put a dot on the rim and let it walk anticlockwise. The amount it has turned, measured from the positive $x$ -axis, is the angle $θ$ .

WHY this picture. Every trig equation is secretly a question about this walking dot. If we can see the dot's position, we can see every solution at once.

PICTURE. The dot sits at coordinates $(cos θ, sin θ)$ .

Figure — Solving trig equations — general solutions, solutions in given range

So " $sin θ = s$ " literally asks: at which angles is the dot exactly $s$ high?

Step 2 — Why there are infinitely many answers

WHAT. Let the dot keep walking past one full loop. After $2 π$ radians it returns to exactly the same spot.

WHY. Same spot $\Rightarrow$ same height $\Rightarrow$ same $sin$ , same $cos$ . So if angle $θ_{0}$ is a solution, so is $θ_{0} + 2 π$ , and $θ_{0} + 4 π$ , and so on forever. This repetition is called periodicity.

PICTURE. Three dots stacked at the same rim position, labelled $θ_{0}$ , $θ_{0} + 2 π$ , $θ_{0} + 4 π$ — same place, different lap.

Step 3 — Cosine: why the answer is $2 nπ \pm α$

WHAT. Solve $cos θ = c$ . Draw the vertical line $x = c$ . Where it crosses the circle, the dot has horizontal coordinate exactly $c$ — those crossings are the solutions.

WHY. $cos θ$ is the horizontal coordinate. Fixing it to $c$ is the same as fixing the vertical line $x = c$ . A vertical line crosses the circle in two points, and those two points are mirror images across the $x$ -axis.

PICTURE. The line $x = c$ hits the circle at height $+$ and height $-$ : angles $+ α$ and $- α$ .

The upper crossing is the principal value $α = cos^{- 1} c$ (see Inverse Trigonometric Functions). The lower crossing is its mirror, at angle $- α$ — because reflecting a point across the $x$ -axis flips the height but keeps the horizontal coordinate. That mirror symmetry is exactly the statement $cos (- θ) = cos θ$ (cosine is even).

Now attach the "which lap" term from Step 2:

Step 4 — Sine: why the answer is $nπ + (- 1)^{n} α$

WHAT. Solve $sin θ = s$ . Now draw a horizontal line $y = s$ . Its crossings with the circle are the solutions.

WHY. $sin θ$ is the height. Fixing height $= s$ is fixing the horizontal line. It crosses the circle in two points that are mirror images across the vertical line $x = 0$ (the $y$ -axis).

PICTURE. The line $y = s$ hits at angle $α$ on the right and at angle $π - α$ on the left.

Reflecting a point across the y-axis keeps its height but sends angle $α$ to $π - α$ . So the two solutions per lap are $α$ and $π - α$ — they are supplementary, NOT $\pm α$ . (This is the single most common sine mistake.)

Why the strange $(- 1)^{n}$ . We want one tidy formula that spits out $α, π - α, 2 π + α, 3 π - α, \dots$ as $n = 0, 1, 2, 3, \dots$ . Watch the pattern:

$n$	$nπ + (- 1)^{n} α$	value
$0$	$0 + (+ 1) α$	$α$
$1$	$π + (- 1) α$	$π - α$
$2$	$2 π + (+ 1) α$	$2 π + α$
$3$	$3 π + (- 1) α$	$3 π - α$

PICTURE (the alternation). As $n$ climbs, the dot alternates between the two mirror families, one per half-turn.

Step 5 — Tangent: why the answer is $nπ + α$ (just one family)

WHAT. Solve $tan θ = t$ . Here $tan θ = \frac{sin θ}{cos θ} = \frac{height}{horizontal}$ — the slope of the line from the origin to the dot.

WHY only one family. A slope is a direction of a line. But a line through the origin has the dot pointing one way and the exact opposite way — half a turn ( $π$ ) apart — with the same slope. So $tan$ repeats every $π$ , not every $2 π$ . Within each half-turn it hits every slope exactly once.

PICTURE. One straight line through the origin at slope $t$ ; the two crossings with the circle are $α$ and $α + π$ — but they share one slope, so we only need one base angle plus steps of $π$ .

Step 6 — The degenerate / edge cases (never skip these)

WHAT. What happens at the special values where a line is tangent to the circle, or misses it?

WHY. These are exactly the cases where "two solutions per lap" quietly becomes one or zero. If you don't know them, a real exam question will trap you.

PICTURE. Four panels: cosine peak, sine peak, the miss, and the tan vertical case.

The one-picture summary

WHAT. Cosine reads the horizontal coordinate (vertical cutting line → mirror across $x$ -axis → $\pm α$ ). Sine reads the height (horizontal cutting line → mirror across $y$ -axis → $α, π - α$ ). Tangent reads the slope (a line through the origin → one direction repeats every $π$ ). The lap term ( $2 nπ$ or $nπ$ ) stacks copies forever.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a dot walking round a circle of radius $1$ forever. Its shadow on the ground (left–right) is $cos$ ; its height is $sin$ ; the tilt of the string from the centre to the dot is $tan$ .

Ask "when is the shadow at position $c$ ?" — draw an up-down line there; it clips the circle at two spots that mirror over the flat ground line, one above, one below. Those are $+ α$ and $- α$ . Every extra loop repeats them: that's $2 nπ \pm α$ .

Ask "when is the dot at height $s$ ?" — draw a flat line there; it clips two spots that mirror over the up-down line, one on the right ( $α$ ) and one on the left ( $π - α$ ). Feed $n = 0, 1, 2, \dots$ into $nπ + (- 1)^{n} α$ and it flips politely between those two, forever.

Ask "when is the string at slope $t$ ?" — a string and its exact opposite point the same slope, half a turn apart, so you only ever need one angle repeated every $π$ : $nπ + α$ .

And if the line you draw never reaches the circle (height bigger than $1$ , shadow bigger than $1$ ), there simply is no dot there — the honest answer is no solution.

Recall Quick self-test

Why does cosine get $\pm α$ but sine gets $α, π - α$ ? ::: Cosine fixes the horizontal line so the crossings mirror over the $x$ -axis (giving $\pm α$ ); sine fixes the horizontal cutting line at height s so crossings mirror over the $y$ -axis (giving $α$ and $π - α$ ). Why does tan have period $π$ not $2 π$ ? ::: A line through the origin points in two opposite directions half a turn apart with the same slope, so $tan$ repeats every $π$ .

Connections

Unit Circle and Radian Measure — the walking-dot picture behind every step.
Graphs of Trigonometric Functions — same repetition, seen as a wave.
Inverse Trigonometric Functions — where the principal $α$ comes from.
Trigonometric Identities — $tan = sin / cos$ used in Step 5.
Compound and Double Angle Formulae — needed once the angle is doubled.
Hinglish parent note