1. Why trig equations are different
A polynomial equation like $2x - 5 = 0$ has one answer. A quadratic has at most two. The number of solutions is bounded by the degree, and once you find them you are done.
Trig equations break that pattern. Consider $\sin x = \tfrac{1}{2}$. The first angle that comes to mind is $x = \pi/6$. But $\sin(5\pi/6) = \tfrac{1}{2}$ as well. And $\sin(\pi/6 + 2\pi) = \tfrac{1}{2}$. And $\sin(5\pi/6 + 2\pi)$, and $\sin(\pi/6 - 2\pi)$, and on forever. The sine wave crosses the height $\tfrac{1}{2}$ twice per cycle, and there are infinitely many cycles.
Because $\sin$ and $\cos$ have period $2\pi$ and $\tan$ has period $\pi$, every trig equation with even one solution has infinitely many. Reporting "the" answer means reporting all of them — either as a finite list inside one period, or as a formula in some integer $k$ that runs over $\mathbb{Z}$.
So the very first question to settle, before any algebra, is: what does "solve" mean here? Two phrasings dominate.
- Report every solution that sits in one full revolution.
- Finite answer — usually 1, 2, 3, or 4 values.
- What "find all solutions on $[0, 2\pi)$" or "find all angles between $0$ and $360°$" means.
- Report the general solution with $+ 2\pi k$ (or $+ \pi k$ for tan).
- $k$ ranges over $\mathbb{Z}$ — every integer, positive, negative, zero.
- What "find the general solution" or "solve over $\mathbb{R}$" means.
Both phrasings ask for the same set of angles; they just describe it differently. The first names the solutions one by one, the second packages them into a periodic formula.
2. The three basic equations
Almost every trig equation eventually reduces, through algebra and identities, to one of three building blocks:
$$ \sin x = c, \quad \cos x = c, \quad \tan x = c $$Feasibility comes first. For $\sin$ and $\cos$ the output is always in $[-1, 1]$, so if $|c| > 1$ there is no solution at all — done. For $\tan$, $c$ can be any real number; there is always a solution.
For a basic equation $\sin x = c$ (or $\cos$, or $\tan$), the reference angle is $\arcsin |c|$ (resp. $\arccos |c|$ or $\arctan |c|$) — the acute, positive angle whose sine, cosine, or tangent has the same magnitude as $c$. Every solution is the reference angle reflected into the appropriate quadrants by sign.
The unit circle does the rest. Each trig function is positive in two quadrants and negative in two. The reference angle picks one of four spots in each quadrant; the sign of $c$ tells you which two are right.
| Function | Positive in | Negative in | Period |
|---|---|---|---|
| $\sin$ | Q I, Q II | Q III, Q IV | $2\pi$ |
| $\cos$ | Q I, Q IV | Q II, Q III | $2\pi$ |
| $\tan$ | Q I, Q III | Q II, Q IV | $\pi$ |
Worked walkthrough for $\sin x = \tfrac{1}{2}$ on $[0, 2\pi)$. The reference angle is $\arcsin\tfrac{1}{2} = \pi/6$. Sine is positive, so we want Q I and Q II:
- Quadrant I: the reference angle itself — $x = \pi/6$.
- Quadrant II: $\pi$ minus the reference — $x = \pi - \pi/6 = 5\pi/6$.
So $x \in \{\pi/6,\ 5\pi/6\}$. The same recipe works for cosine (Q I gives the reference, Q IV gives $2\pi$ minus it; flip to Q II / Q III when $c$ is negative) and for tangent (Q I gives the reference, Q III adds $\pi$).
3. General vs. principal solutions
The principal solutions are the ones inside $[0, 2\pi)$ — the finite list. The general solution wraps that list around the whole real line by adding integer multiples of the period.
For sine and cosine, every period of length $2\pi$ contains the same two solutions, so we add $+\,2\pi k$. Sine in particular has a cleaner phrasing using its reflection symmetry around $\pi/2$:
$$ \sin x = c \quad\Longleftrightarrow\quad x = \arcsin c + 2\pi k \ \text{ or } \ x = \pi - \arcsin c + 2\pi k, \quad k \in \mathbb{Z} $$Cosine has reflection symmetry around the x-axis, which lets us collapse its two solutions into a single $\pm$:
$$ \cos x = c \quad\Longleftrightarrow\quad x = \pm\arccos c + 2\pi k, \quad k \in \mathbb{Z} $$Tangent is the odd one out. It repeats every $\pi$ — not every $2\pi$ — because $\tan(x + \pi) = \tan x$. Within a single period of length $\pi$, $\tan x = c$ has exactly one solution. So the general form is:
$$ \tan x = c \quad\Longleftrightarrow\quad x = \arctan c + \pi k, \quad k \in \mathbb{Z} $$Using $+\,2\pi k$ for a tangent equation. Tan's period is $\pi$, so $+\,2\pi k$ misses every other solution. If you wrote $\tan x = 1 \Rightarrow x = \pi/4 + 2\pi k$, you forgot $x = 5\pi/4$ — and infinitely many of its cousins.
A few values worth memorising
| Equation | Principal solutions on $[0, 2\pi)$ | General solution |
|---|---|---|
| $\sin x = 0$ | $0,\ \pi$ | $x = \pi k$ |
| $\sin x = 1$ | $\pi/2$ | $x = \pi/2 + 2\pi k$ |
| $\sin x = -1$ | $3\pi/2$ | $x = -\pi/2 + 2\pi k$ |
| $\cos x = 0$ | $\pi/2,\ 3\pi/2$ | $x = \pi/2 + \pi k$ |
| $\cos x = 1$ | $0$ | $x = 2\pi k$ |
| $\cos x = -1$ | $\pi$ | $x = \pi + 2\pi k$ |
| $\tan x = 0$ | $0,\ \pi$ | $x = \pi k$ |
| $\tan x = 1$ | $\pi/4,\ 5\pi/4$ | $x = \pi/4 + \pi k$ |
4. A solving strategy
Most trig equations are not handed to you in basic form. The work is reducing them. A reliable sequence:
- Isolate a single trig expression on one side, or move everything to one side equal to zero (so you can factor).
- Reduce to one function using identities — Pythagorean ($\sin^2 + \cos^2 = 1$), double-angle ($\sin 2x = 2\sin x\cos x$, $\cos 2x = 2\cos^2 x - 1$), etc.
- Substitute $u = \sin x$ (or $\cos x$, or $\tan x$) if what you have is a polynomial in one trig function.
- Factor if it's a product equal to zero, then set each factor to zero.
- Solve each basic equation for $x$ — in the requested range, or as a general solution.
- Verify if you squared, divided by an expression involving an unknown, or used another non-reversible step.
Not every step is needed every time. A simple $\cos x = -\tfrac{\sqrt{2}}{2}$ skips straight to step 5. A grim $\sin 2x + \cos 2x = 1$ might use all six.
5. Four techniques
Substitution: when the equation is a polynomial in one trig function
If the only trig expression appearing is, say, $\sin x$ — and it appears squared, cubed, or in a linear combination — let $u = \sin x$ and solve the resulting polynomial. Each root $u_i$ becomes its own basic equation $\sin x = u_i$.
Example: $2\sin^2 x - \sin x - 1 = 0$. Let $u = \sin x$:
$$ 2u^2 - u - 1 = 0 \quad\Longrightarrow\quad (2u + 1)(u - 1) = 0 \quad\Longrightarrow\quad u = -\tfrac{1}{2}\ \text{or}\ u = 1 $$Now solve each piece. $\sin x = 1$ gives $x = \pi/2$. $\sin x = -\tfrac{1}{2}$ gives $x = 7\pi/6$ or $11\pi/6$. Three solutions on $[0, 2\pi)$.
Identity reduction: get everything in one function
If both $\sin$ and $\cos$ appear, use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to rewrite one in terms of the other, then substitute. Double-angle identities do the same job when $\sin 2x$ or $\cos 2x$ appears alongside $\sin x$ or $\cos x$.
Example: $\cos 2x + \cos x = 0$. Replace $\cos 2x = 2\cos^2 x - 1$:
$$ 2\cos^2 x + \cos x - 1 = 0 $$Let $u = \cos x$: $(2u - 1)(u + 1) = 0$, so $u = \tfrac{1}{2}$ or $u = -1$. Solutions on $[0, 2\pi)$: $x = \pi/3,\ \pi,\ 5\pi/3$.
Factoring: a product equal to zero
If you can rewrite the equation as a product of trig expressions equal to zero, set each factor to zero and solve separately. The trick is to not divide away factors — you would lose solutions.
Example: $\sin 2x = \sin x$. Move everything to one side and use $\sin 2x = 2\sin x\cos x$:
$$ 2\sin x\cos x - \sin x = 0 \quad\Longrightarrow\quad \sin x(2\cos x - 1) = 0 $$So $\sin x = 0$ ($x = 0,\ \pi$) or $\cos x = \tfrac{1}{2}$ ($x = \pi/3,\ 5\pi/3$). Four solutions on $[0, 2\pi)$.
A natural-looking move on $\sin 2x = \sin x$ is to "cancel" $\sin x$ from both sides, getting $2\cos x = 1$. That throws away every solution where $\sin x = 0$ — including $x = 0$ and $x = \pi$, both of which satisfy the original. Whenever you are tempted to divide by an expression that contains the unknown, move it to the other side and factor instead.
Squaring: powerful but introduces extraneous solutions
If $\sin x$ and $\cos x$ appear in a stubborn combination like $\sin x + \cos x = 1$, squaring both sides collapses them via the Pythagorean identity. The cost: squaring is not reversible, so some "solutions" of the squared equation will not satisfy the original. You must check every candidate.
Example: $\sin x + \cos x = 1$ on $[0, 2\pi)$. Square:
$$ (\sin x + \cos x)^2 = 1 \quad\Longrightarrow\quad 1 + 2\sin x \cos x = 1 \quad\Longrightarrow\quad \sin 2x = 0 $$So $2x = \pi k$, giving candidates $x = 0,\ \pi/2,\ \pi,\ 3\pi/2$. Check each in the original equation:
| $x$ | $\sin x + \cos x$ | Keep? |
|---|---|---|
| $0$ | $0 + 1 = 1$ | Yes |
| $\pi/2$ | $1 + 0 = 1$ | Yes |
| $\pi$ | $0 + (-1) = -1$ | No — extraneous |
| $3\pi/2$ | $-1 + 0 = -1$ | No — extraneous |
Real solutions: $x = 0,\ \pi/2$. Two of the four candidates were ghosts produced by the squaring step — they satisfied $(\sin x + \cos x)^2 = 1$ (because $(-1)^2 = 1$ too) but not the unsquared equation.
6. Multiple-angle equations
An equation like $\sin 2x = \tfrac{1}{2}$ or $\tan 3x = 1$ has the unknown wrapped inside a scaling. The clean way to handle it: solve for the wrapped quantity first, then divide.
For $\sin 2x = \tfrac{1}{2}$ on $[0, 2\pi)$, let $u = 2x$. As $x$ ranges over $[0, 2\pi)$, $u$ ranges over $[0, 4\pi)$ — two full periods. The solutions of $\sin u = \tfrac{1}{2}$ in that doubled range are:
$$ u = \pi/6,\ 5\pi/6,\ 13\pi/6,\ 17\pi/6 $$Dividing by 2:
$$ x = \pi/12,\ 5\pi/12,\ 13\pi/12,\ 17\pi/12 $$Four solutions, not two. Doubling the angle inside packs two periods into the original window, so you get twice as many crossings.
For $\sin(kx) = c$ or $\cos(kx) = c$ on $[0, 2\pi)$ with $|c| < 1$, expect $2k$ solutions. For $\tan(kx) = c$ on $[0, 2\pi)$, expect $2k$ solutions as well (tan repeats $2k$ times in a $2\pi$-window). If your count is off, you probably divided too early or forgot to extend the auxiliary range.
General solution form: solve $\sin(kx) = c$ as $kx = \arcsin c + 2\pi n$ or $kx = \pi - \arcsin c + 2\pi n$, then divide everything — including the period term — by $k$:
$$ x = \frac{\arcsin c}{k} + \frac{2\pi n}{k}, \quad x = \frac{\pi - \arcsin c}{k} + \frac{2\pi n}{k} $$The new "period in $x$" is $2\pi / k$ for sine and cosine, or $\pi / k$ for tangent. Forgetting to divide the period is another classic error.
7. Common pitfalls
$\arcsin\tfrac{1}{2} = \pi/6$ is a real solution of $\sin x = \tfrac{1}{2}$, but it is one of infinitely many. On $[0, 2\pi)$ alone there is also $x = 5\pi/6$. If the question asks "find all solutions" and you write $x = \pi/6$, you have not answered.
Tan's period is $\pi$. The general solution of $\tan x = c$ is $x = \arctan c + \pi k$, not $+ 2\pi k$. Using the wrong period halves the solution set.
Given $\sin 2x = \sin x$, dividing both sides by $\sin x$ to get $2\cos x = 1$ silently throws away the solutions where $\sin x = 0$. Always factor rather than divide.
Squaring both sides is sometimes necessary to combine $\sin$ and $\cos$, but it creates phantom solutions. Every candidate must be substituted into the original equation; the ones that fail are extraneous and are discarded.
$\tan x$ is undefined where $\cos x = 0$; $\cot x$ and $\csc x$ are undefined where $\sin x = 0$. If a candidate makes the original equation involve $\infty$, reject it — it is not a solution.
When solving $\sin 2x = c$ on $[0, 2\pi)$, the auxiliary variable $u = 2x$ runs over $[0, 4\pi)$ — two periods. If you only look for solutions in $[0, 2\pi)$, you miss half. The rule of thumb: scale the search range to match the scaling inside.
8. Worked examples
In rough order of difficulty. Try each yourself before opening the solution.
Example 1 · Basic — solve $\cos x = -\tfrac{\sqrt{2}}{2}$ on $[0, 2\pi)$
Step 1. Reference angle: $\arccos\tfrac{\sqrt{2}}{2} = \pi/4$.
Step 2. Cosine is negative, so the solutions live in Q II and Q III.
Step 3. Q II: $\pi - \pi/4 = 3\pi/4$. Q III: $\pi + \pi/4 = 5\pi/4$.
Answer. $x = 3\pi/4,\ 5\pi/4$.
Example 2 · Tan, general solution — solve $\tan x = \sqrt{3}$
Step 1. $\arctan\sqrt{3} = \pi/3$.
Step 2. Tan has period $\pi$, so the general solution adds $\pi k$:
$$ x = \pi/3 + \pi k, \quad k \in \mathbb{Z} $$Check. On $[0, 2\pi)$ this gives $x = \pi/3$ (Q I) and $x = 4\pi/3$ (Q III) — both with $\tan = \sqrt{3}$, as expected from the period-$\pi$ pattern.
Example 3 · Quadratic substitution — solve $2\sin^2 x - \sin x - 1 = 0$ on $[0, 2\pi)$
Step 1. Let $u = \sin x$:
$$ 2u^2 - u - 1 = 0 \quad\Longrightarrow\quad (2u + 1)(u - 1) = 0 $$So $u = -\tfrac{1}{2}$ or $u = 1$.
Step 2. $\sin x = 1$ gives $x = \pi/2$ (only Q I/II boundary, just one solution).
Step 3. $\sin x = -\tfrac{1}{2}$: reference $\pi/6$, sine negative in Q III and Q IV, so $x = \pi + \pi/6 = 7\pi/6$ and $x = 2\pi - \pi/6 = 11\pi/6$.
Answer. $x = \pi/2,\ 7\pi/6,\ 11\pi/6$.
Example 4 · Identity reduction — solve $\cos 2x + \cos x = 0$ on $[0, 2\pi)$
Step 1. Replace $\cos 2x = 2\cos^2 x - 1$:
$$ 2\cos^2 x - 1 + \cos x = 0 $$Step 2. Let $u = \cos x$: $2u^2 + u - 1 = 0 \Rightarrow (2u - 1)(u + 1) = 0$. So $u = \tfrac{1}{2}$ or $u = -1$.
Step 3. $\cos x = \tfrac{1}{2}$ gives $x = \pi/3,\ 5\pi/3$. $\cos x = -1$ gives $x = \pi$.
Answer. $x = \pi/3,\ \pi,\ 5\pi/3$.
Example 5 · Factoring, don't divide — solve $\sin 2x = \sin x$ on $[0, 2\pi)$
Step 1. Use $\sin 2x = 2\sin x \cos x$ and move everything to one side:
$$ 2\sin x \cos x - \sin x = 0 \quad\Longrightarrow\quad \sin x(2\cos x - 1) = 0 $$Step 2. $\sin x = 0$: $x = 0,\ \pi$.
Step 3. $\cos x = \tfrac{1}{2}$: $x = \pi/3,\ 5\pi/3$.
Answer. $x = 0,\ \pi/3,\ \pi,\ 5\pi/3$. (Cancelling $\sin x$ at step 1 would have lost $x = 0$ and $x = \pi$.)
Example 6 · Multiple angle — solve $\tan 2x = 1$ on $[0, 2\pi)$
Step 1. Let $u = 2x$. As $x$ runs over $[0, 2\pi)$, $u$ runs over $[0, 4\pi)$.
Step 2. $\tan u = 1$ has solutions $u = \pi/4 + \pi k$. In $[0, 4\pi)$ that means $k = 0, 1, 2, 3$:
$$ u = \pi/4,\ 5\pi/4,\ 9\pi/4,\ 13\pi/4 $$Step 3. Divide each by 2:
$$ x = \pi/8,\ 5\pi/8,\ 9\pi/8,\ 13\pi/8 $$Answer. Four solutions.
Example 7 · Squaring with extraneous check — solve $\sin x = \cos x$ on $[0, 2\pi)$
Approach A (preferred). Divide both sides by $\cos x$ — this is safe because $\cos x = 0$ would force $\sin x = 0$ too, which never happens simultaneously. Result: $\tan x = 1$, so $x = \pi/4,\ 5\pi/4$.
Approach B (squaring, for practice). Square both sides: $\sin^2 x = \cos^2 x \Rightarrow \sin^2 x - \cos^2 x = 0 \Rightarrow -\cos 2x = 0 \Rightarrow \cos 2x = 0$.
$2x = \pi/2 + \pi k$, so $x = \pi/4 + \pi k/2$, giving candidates $\pi/4,\ 3\pi/4,\ 5\pi/4,\ 7\pi/4$.
Check in the original $\sin x = \cos x$:
- $\pi/4$: $\tfrac{\sqrt{2}}{2} = \tfrac{\sqrt{2}}{2}$ ✓
- $3\pi/4$: $\tfrac{\sqrt{2}}{2} \neq -\tfrac{\sqrt{2}}{2}$ ✗ (extraneous)
- $5\pi/4$: $-\tfrac{\sqrt{2}}{2} = -\tfrac{\sqrt{2}}{2}$ ✓
- $7\pi/4$: $-\tfrac{\sqrt{2}}{2} \neq \tfrac{\sqrt{2}}{2}$ ✗ (extraneous)
Answer. $x = \pi/4,\ 5\pi/4$. Both approaches agree, but squaring required a check.
Example 8 · Squaring with extraneous check — solve $\sin x + \cos x = 1$ on $[0, 2\pi)$
Step 1. Square both sides:
$$ (\sin x + \cos x)^2 = 1 \quad\Longrightarrow\quad 1 + 2\sin x \cos x = 1 \quad\Longrightarrow\quad \sin 2x = 0 $$Step 2. $2x = \pi k$, giving candidates $x = 0,\ \pi/2,\ \pi,\ 3\pi/2$.
Step 3. Verify in the original:
- $x = 0$: $0 + 1 = 1$ ✓
- $x = \pi/2$: $1 + 0 = 1$ ✓
- $x = \pi$: $0 + (-1) = -1$ ✗
- $x = 3\pi/2$: $-1 + 0 = -1$ ✗
Answer. $x = 0,\ \pi/2$.
Example 9 · No solution — solve $\sin x = 2$
$|\sin x| \leq 1$ for every real $x$, and $2 > 1$. No solution. Recognising infeasibility is part of the work — if $|c| > 1$ for a sine or cosine equation, stop immediately.