A calculator gives arcsin(1/2)=π/6. Why does a signal cross that level again? An inverse returns one principal angle; an equation needs every angle. Use equation transformations, inverses and identities.

Choosing an inverse branch

An inverse asks which angle has a given sine, cosine, tangent or cotangent. Restrict the original domain to make the answer unique. For example, arcsin a is the angle in [−π/2,π/2] whose sine equals a. Principal-value conventions:

arcsin⁡a ∈ [ − π/2,π/2], arccos⁡a ∈ [0,π] (∣a∣ ≤ 1).

arctan⁡a ∈ ( − π/2,π/2), arccot⁡a ∈ (0,π) ( − ∞ < a < ∞).

Our arccot uses (0,π), matching the local textbook; another convention needs conversion. Examples: arccos(−1/2)=2π/3, arctan1=π/4, arccot0=π/2. Calculator sin⁻¹ means arcsin, not 1/sin.

All solutions in radians

In these formulas k is any integer:

sin⁡x = a: x = arcsin⁡a + 2πk or x = π − arcsin⁡a + 2πk.

cos⁡x = a: x = ± arccos⁡a + 2πk;

tan⁡x = a: x = arctan⁡a + πk; cot⁡x = a: x = arccot⁡a + πk.

Equal circle heights give x and π−x, hence two sine families; cosine agrees at x and −x. A full turn adds 2π; tan/cot repeat after π. For sin/cos, |a|>1 gives no real solutions. At a=±1 some families coincide; do not count them twice. OpenStax

One level, several times

sin(πt/3)=1/2

036912-101t, hy
Solid sin(πt/3), dashed level 1/2. All four crossings on 0≤t<12: 0.5,2.5,6.5,8.5 h. Uniform scales, dimensionless output.

For a hypothetical 6-hour signal:

sin⁡(πt/3) = 1/2 ⟺ t = 1/2 + 6k or t = 5/2 + 6k.

On 0≤t<12, retain 0.5,2.5,6.5,8.5 hours. Apply time limits after finding all solutions.

Transform without losing angles

sin⁡2x = sin⁡x ⟺ sin⁡x(2cos⁡x − 1) = 0.

Thus x=πk or x=±π/3+2πk. Dividing by sinx loses the first family. For sin²x−3sinx+2=0, substitution has a restricted range:

(u)^(2) − 3u + 2 = 0, u = sin⁡x ∈ [ − 1,1] ⇒ u = 1 ⇒ x = π/2 + 2πk.

u=2 is excluded. Preserve original denominators and substituted ranges; check against the original signal, not only the transformed polynomial.