Every trigonometric equation is eventually reduced to the basic ones (sin x = a, cos x = a, tg x = a). So learn those perfectly first, then the reduction techniques. Because of periodicity there are infinitely many roots, written with a parameter n ∈ Z.
§1. Basic equations
sin x = a (|a| ≤ 1): x = (−1)ⁿ arcsin a + πn
cos x = a (|a| ≤ 1): x = ± arccos a + 2πn
tg x = a (any a): x = arctg a + πn
For |a| > 1, sin x = a and cos x = a have no solutions
Special cases (simpler than the general formula):
sin x = 0 → x = πn; sin x = 1 → x = π/2 + 2πn; sin x = −1 → x = −π/2 + 2πn.
cos x = 0 → x = π/2 + πn; cos x = 1 → x = 2πn; cos x = −1 → x = π + 2πn.
Inverse functions:arcsin a ∈ [−π/2; π/2], arccos a ∈ [0; π], arctg a ∈ (−π/2; π/2). arcsin(−a) = −arcsin a, arccos(−a) = π − arccos a.
Worked example
cos x = −1/2. arccos(−1/2) = π − π/3 = 2π/3 → x = ±2π/3 + 2πn.
§2. One function and substitution
Rewrite the equation in one function and set t = sin x (or cos x, tg x). Remember |t| ≤ 1.
Worked example
2cos²x + sin x − 1 = 0.
cos²x = 1 − sin²x: 2 − 2sin²x + sin x − 1 = 0 → 2sin²x − sin x − 1 = 0.
t = sin x: 2t² − t − 1 = 0 → t = 1 or t = −1/2.
sin x = 1 → x = π/2 + 2πn; sin x = −1/2 → x = (−1)ⁿ⁺¹ π/6 + πn.
§3. Homogeneous equations
a sin x + b cos x = 0 — divide by cos x (after checking cos x = 0 isn't a root): tg x = −b/a. a sin²x + b sin x cos x + c cos²x = 0 — divide by cos²x: a tg²x + b tg x + c = 0.
Worked example
sin²x − 3 sin x cos x + 2cos²x = 0.
If cos x = 0, then sin²x = 0 too — impossible, so dividing is safe.
tg²x − 3tg x + 2 = 0 → tg x = 1 or tg x = 2.
x = π/4 + πn, x = arctg 2 + πn.
§4. Factoring and formulas
Never divide both sides by an expression containing the variable — you lose roots. Factor out instead.
Worked example
sin 2x = cos x.
2 sin x cos x − cos x = 0 → cos x (2 sin x − 1) = 0.
cos x = 0 → x = π/2 + πn; sin x = 1/2 → x = (−1)ⁿ π/6 + πn.
Worked example (from an official sample)
sin 7x · cos x = sin 6x. How many roots lie in [−π; π]?
Product to sum: ½(sin 8x + sin 6x) = sin 6x → sin 8x = sin 6x.
8x = 6x + 2πk → x = πk; or 8x = π − 6x + 2πk → x = π/14 + πk/7 = π(2k + 1)/14.
The first series gives −π, 0, π — 3 roots; the second gives odd 2k + 1 from −13 to 13 — 14 roots; the series don't overlap.
Total 17; the smallest positive root is π/14.
§5. a sin x + b cos x = c — auxiliary angle
Divide both sides by √(a² + b²). If |c| > √(a² + b²), there are no solutions.
Worked example
sin x + cos x = 1 → √2 sin(x + π/4) = 1 → sin(x + π/4) = √2/2 → x = 2πn or x = π/2 + 2πn.
§6. Selecting roots in a given interval
Many test questions ask not for the general solution but for the roots in a given interval, their number or their sum. The general solution produces infinitely many numbers, so the ones lying in the interval have to be picked out. There are three ways to do it:
Trying integer values of `n`. Substitute n = 0, ±1, ±2, … into the general solution and keep going until you leave the interval. The most reliable and simplest method.
Using an inequality. Solve the double inequality a ≤ x₀ + πn ≤ b for n and find the suitable integers. Convenient when the interval is long.
On the unit circle. Mark the roots on the circle, highlight the arc corresponding to the interval and count the points on it. Good for seeing the number of roots quickly.
Check the endpoints separately: boundary points such as x = 0, x = π, x = 2π belong to a closed interval [a; b] but not to an open interval (a; b).
When a solution has two families (for example, the + and − families for cos x = a, or two equations after factorising), check each family separately and count any repeated root only once.
Worked example
Find the roots of cos 2x = 1/2 in the interval [0; π].
2x = ±π/3 + 2πn → x = ±π/6 + πn.
Family x = π/6 + πn: n = 0 gives π/6 (inside), n = 1 gives 7π/6 > π (outside).
Family x = −π/6 + πn: n = 0 gives −π/6 < 0 (outside), n = 1 gives 5π/6 (inside).
Check: cos(π/3) = 1/2, cos(5π/3) = 1/2 — correct.
Answer: π/6 and 5π/6 — two roots.
Check yourself
tg x = −1.
sin x = 2.
cos 2x = 0.
How many roots does sin x = 1/2 have on [0; 2π]?
Smallest positive root of sin x cos x = 1/4.
sin x − √3 cos x = 0.
Find the roots of tg x = √3 in [0; 2π].
How many roots does 2sin²x − sin x = 0 have in [0; 2π]?
Common mistakes
Dividing both sides by sin x or cos x and losing roots.
Writing arcsin a when |a| > 1.
Forgetting ± for cos x = a and (−1)ⁿ for sin x = a.
Solving sin 2x = 1/2 but not halving the πn term too.
Forgetting the endpoints (0, 2π) when counting roots in an interval, or counting a root shared by two families twice.