Trigonometric Inequalities
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Direct answer
Pick the interval before touching the algebra: a trigonometric inequality is solved on a declared domain such as [0, 2π), and the answer is a union of sub-intervals, never isolated values. The engine is the unit circle read on its monotonic halves: sin x > sin α gives x ∈ (α, π − α) per period; cos x > cos α gives (−α, α); tan x > tan α gives (α, π/2) per period — the jump at π/2 splits every tangent inequality across its asymptote. Compound expressions reduce first to a single wave, A sin x + B cos x = R sin(x + φ) with R = √(A² + B²), after which only a plain sine inequality remains.
What you must remember
- Sine: sin x > sin α on [0, 2π) gives (α, π − α); sin x < sin α gives the complement pieces (0, α) ∪ (π − α, 2π).
- Cosine: cos x > cos α gives (0, α) ∪ (2π − α, 2π) — the band around 0, since cosine peaks at the origin of each period.
- Tangent: tan x > tan α gives (α, π/2) ∪ (α + π, 3π/2); the asymptote at π/2 always splits the set.
- Single-wave reduction: A sin x + B cos x = R sin(x + φ) with R = √(A² + B²) and tan φ = B/A — then solve sin(x + φ) > c and shift back.
- Graph discipline: sketch one period before writing the answer; the graph of sin x above the line y = 1/2 is two vertical strips, not two points.
- Equality endpoints: with ≥ or ≤, endpoints join the solution; JEE numerical answers hinge on the bracket type.
- Substitution route: for quadratics in sin x, set t = sin x ∈ [−1, 1], solve with the wavy curve, then return to x — the range restriction is where marks are won.
Solving on the unit circle
Solve sin x + cos x ≥ 1 on [0, 2π). Write the left side as √2 sin(x + π/4), so the inequality becomes sin(x + π/4) ≥ 1/√2. One round of the circle gives π/4 ≤ x + π/4 ≤ 3π/4, hence x ∈ [0, π/2] — a clean quarter-period answer. Notice the method's economy: no squaring, no casework, one substitution. Contrast a tangent problem: solve tan x ≥ 1 on [0, 2π). The tempting answer (π/4, π/2] is incomplete because tangent repeats after π, and on [π, 3π/2) it again climbs from 0 to +∞; the full solution is [π/4, π/2) ∪ [5π/4, 3π/2) — and note π/2 itself is excluded both times, since tan x has no value there. The unit circle picture (tangent = slope of the terminal ray) makes the split automatic: the ray is steeper than 45° in the first and third quadrants only, stopping short of the vertical.
Trap territory in intervals
JEE Main grades the interval writing itself: an answer of (π/4, π/2) where [π/4, π/2) is correct loses the numerical mark, and the option list is built to catch exactly that bracket. Domains like [0, 2π) versus (0, 2π] shift which endpoints are even available. JEE Advanced layers parameters and compositions: solve sin 2x > cos 2x on a restricted interval, or inequalities with |sin x| where the period halves to π — |sin x| ≥ 1/2 removes the middle band of each half-period. Squaring-based attempts on inequalities like sin x + cos x < 1 generate extraneous regions because both sides change sign across the period; the R-substitution route avoids the entire conversation. One rehearsal worth doing before the exam: solve tan x < 1 on [0, 2π) and write the answer with the asymptote gap visible — [0, π/4) ∪ (π/2, 5π/4) ∪ (3π/2, 2π), where the negative stretches of tangent join automatically because everything below zero is certainly below one. Getting every piece of a tan inequality is the single best drill this topic offers.
Frequently asked questions
How do you solve sin x > 1/2 on [0, 2π)?
x ∈ (π/6, 5π/6) — the arc where the unit circle's sine coordinate exceeds 1/2 within one round.
Why must tangent inequalities be split at π/2?
Tangent jumps from +∞ to −∞ at π/2, so the inequality's truth flips at the asymptote; crossing it without splitting fabricates or loses solution pieces.
How does A sin x + B cos x reduce?
It equals R sin(x + φ) with R = √(A² + B²) and tan φ = B/A, converting the compound inequality into a single sine inequality.
When may you square a trigonometric inequality?
Only when both sides are known to be non-negative on the interval considered; otherwise squaring manufactures extraneous regions.
What changes when |sin x| replaces sin x?
The period halves to π and the graph lifts below the axis, so solution sets mirror across the x-axis and the answer doubles its pieces per round.