Trigonometric Equations and Inequalities

The Simplest Equations

Here n is any integer (n ∈ ℤ).

Equation	Solution
sin x = a, −1 ≤ a ≤ 1	x = (−1)ⁿ · arcsin a + πn
cos x = a, −1 ≤ a ≤ 1	x = ± arccos a + 2πn
tg x = a	x = arctg a + πn
ctg x = a	x = arcctg a + πn

• If a > 1 or a < −1, there is no solution (for sin x = a and cos x = a).
• For tg and ctg add +πn, for sin and cos add +2πn.
• Special cases: for sin x = 0, x = πn; for sin x = 1, x = π/2 + 2πn; for sin x = −1, x = −π/2 + 2πn.
• For cos x = 0, x = π/2 + πn; for cos x = 1, x = 2πn; for cos x = −1, x = π + 2πn.

Solution Methods

1. Reduce all functions to a single function
2. If it is a quadratic, substitute t = sin x (or cos x) and check |t| ≤ 1
3. Factorize: set each factor equal to zero
4. Solve using the formulas for the simplest equations

• Using the identity sin²x + cos²x = 1, sin and cos are combined.
• Homogeneous equation: dividing a sin x + b cos x = 0 by cos x gives a tg x + b = 0.
• If the argument is multiplied (sin 2x = 0), first find 2x, then divide by two: x = πn/2.

Inequalities

The unit circle or the graph is used, and the answer is always written with the period added.

• For sin x > 0, x ∈ (2πn; π + 2πn).
• cos x < −1/2: (2π/3; 4π/3) + 2πn.
• For cos x > 1/2, x ∈ (−π/3 + 2πn; π/3 + 2πn).
• Period: 2π for sin and cos, π for tg.

Examples

• 2sin²x − sin x − 1 = 0: t = sin x, 2t² − t − 1 = 0, t = 1 or t = −1/2. On the interval [0; 2π], sin x = 1 gives π/2, and sin x = −1/2 gives 7π/6 and 11π/6. In total there are 3 roots.
• tg x = √3: x = π/3 + πn.