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AdvancedGrade 1025 min50 / 82

Trigonometric equations and inequalities

arcsin, arccos and arctan; the solution formulas for sin x = a, cos x = a and tan x = a and their special cases; equations that reduce to quadratics, factoring, homogeneous equations, using identities, selecting roots in an interval, and the simplest trigonometric inequalities on the unit circle.

Check yourself
In this lesson you will learn
  • Explain what arcsin a, arccos a and arctan a mean and find their table values
  • Solve the simplest equations with the general formulas and in the special cases
  • Reduce equations to the simplest ones by substitution, factoring, homogeneous division and identities, and select the roots in a given interval
  • Solve the simplest trigonometric inequalities on the unit circle

In the lesson “Trigonometric functions and their graphs” we computed Elvin’s height on the observation wheel with the formula h(t) = 30 − 28 cos(πt/15) and got h(10) = 44 m. Now let us turn the question around: when is Elvin 44 m high? We have to solve the equation 30 − 28 cos(πt/15) = 44, that is cos(πt/15) = −1/2. The answer is not one number: the cabin passes this height twice in every turn — on the way up and on the way down — and the wheel keeps turning. A trigonometric equation usually has infinitely many roots, and in this lesson we learn to write all of them with one formula.

We will need the unit circle (“Radian measure, basic identities and reduction formulas”), the addition and double-angle formulas (“Addition, double-angle and half-angle formulas”) and the graphs of the functions. We start with new notation — arcsine, arccosine and arctangent — then solve the simplest equations and those that reduce to them, select roots, and finish with inequalities.

Arcsine, arccosine and arctangent

Take the equation sin x = 0.3. The table has no “nice” angle whose sine is 0.3, but such an angle certainly exists: on the segment [−π/2, π/2] the sine increases from −1 to 1 and takes every value exactly once. So there is exactly one number in this segment whose sine is 0.3. We give it a name — arcsin 0.3 (≈ 0.305 rad ≈ 17.5°). For the cosine the same role is played by the segment [0, π], where the cosine decreases from 1 to −1, and for the tangent by the interval (−π/2, π/2).

Definition
arcsin a

For −1 ≤ a ≤ 1, arcsin a is the number in the segment [−π/2, π/2] whose sine equals a. For example, arcsin(1/2) = π/6, because π/6 ∈ [−π/2, π/2] and sin(π/6) = 1/2.

Definition
arccos a

For −1 ≤ a ≤ 1, arccos a is the number in the segment [0, π] whose cosine equals a. For example, arccos(−1) = π and arccos 0 = π/2.

Definition
arctan a

For any a, arctan a is the number in the interval (−π/2, π/2) whose tangent equals a. In the same way, arccot a is the number in (0, π) whose cotangent is a. For example, arctan 1 = π/4, arctan √3 = π/3 and arccot 0 = π/2.

sin(arcsin a) = a cos(arccos a) = a tan(arctan a) = aarcsin(−a) = −arcsin a arccos(−a) = π − arccos a arctan(−a) = −arctan a
where:
  • a−1 ≤ a ≤ 1 for arcsin and arccos, any number for arctan

arcsin and arctan are odd. arccos is not: the values a and −a correspond to points of the unit circle symmetric about the y-axis, and their angles add up to π.

aarcsin aarccos a
−1−π/2π
−√3/2−π/35π/6
−√2/2−π/43π/4
−1/2−π/62π/3
00π/2
1/2π/6π/3
√2/2π/4π/4
√3/2π/3π/6
1π/20
Table values. For the tangent: arctan 0 = 0, arctan(√3/3) = π/6, arctan 1 = π/4, arctan √3 = π/3; for negative a the sign changes.
Values of the inverse trigonometric functions

Evaluate:
1) arcsin(√2/2)
2) arccos(−√3/2)
3) arctan(−1)
4) arcsin(−1/2)
5) arccos 2

Show solution
1) π/4 ∈ [−π/2, π/2] and sin(π/4) = √2/2, so arcsin(√2/2) = π/4.
2) arccos(−a) = π − arccos a: π − π/6 = 5π/6. Check: 5π/6 ∈ [0, π] and cos(5π/6) = −√3/2.
3) arctan is odd: arctan(−1) = −arctan 1 = −π/4.
4) arcsin(−1/2) = −arcsin(1/2) = −π/6 (not 7π/6: that is not in [−π/2, π/2]).
5) A cosine cannot exceed 1, so arccos 2 has no meaning.
A function of an inverse function

Evaluate:
1) sin(arccos(3/5))
2) tan(arcsin(−5/13))
3) arcsin(sin(5π/6))

Show solution
1) Let α = arccos(3/5): cos α = 3/5 and α ∈ [0, π]. On this segment the sine is not negative: sin α = √(1 − 9/25) = 4/5.
2) α = arcsin(−5/13): sin α = −5/13 and α ∈ [−π/2, π/2], where the cosine is not negative: cos α = √(1 − 25/169) = 12/13. tan α = sin α / cos α = −5/12.
3) sin(5π/6) = 1/2 and arcsin(1/2) = π/6. Careful: the answer is not 5π/6 — the arcsine always gives a value in [−π/2, π/2].

The simplest equations: sin x = a, cos x = a, tan x = a

Start with cos x = a. The cosine is the x-coordinate of a point on the unit circle, so we look for the points whose x-coordinate is a — the points where the vertical line x = a meets the circle. If |a| > 1 the line misses the circle and there are no roots. If |a| < 1 there are two points, symmetric about the x-axis: one at the angle arccos a, the other at −arccos a. Every full turn brings the point back, so 2πn is added to both.

cos x = a, |a| ≤ 1: x = ±arccos a + 2πn, n ∈ ℤ
where:
  • athe given number; for |a| > 1 the equation has no roots
  • nany integer (n ∈ ℤ) — the number of whole turns

Two series of roots: arccos a + 2πn and −arccos a + 2πn — two points symmetric about the x-axis.

Equations cos x = a

Solve:
1) cos x = 1/2
2) cos x = −√2/2
3) 2 cos(x/2) = √3
4) cos x = 0.3
5) cos x = −1.5

Show solution
1) x = ±arccos(1/2) + 2πn = ±π/3 + 2πn, n ∈ ℤ.
2) arccos(−√2/2) = π − π/4 = 3π/4, so x = ±3π/4 + 2πn.
3) cos(x/2) = √3/2 ⇒ x/2 = ±π/6 + 2πn. Multiply both sides by 2, including 2πn: x = ±π/3 + 4πn.
4) There is no table value, so the answer is written with the arccosine: x = ±arccos 0.3 + 2πn (arccos 0.3 ≈ 1.266).
5) |−1.5| > 1 — no roots.

In sin x = a the sine is the y-coordinate, so we draw the horizontal line y = a. If |a| < 1 it cuts the circle at two points symmetric about the y-axis: one at the angle arcsin a, the other at π − arcsin a. This gives two series: x = arcsin a + 2πn and x = π − arcsin a + 2πn. They can be combined into one formula: for even n we have (−1)ⁿ = 1 and get the first series, for odd n (−1)ⁿ = −1 and we get the second.

sin x = a, |a| ≤ 1: x = (−1)ⁿ · arcsin a + πn, n ∈ ℤ
where:
  • (−1)ⁿ1 for even n, −1 for odd n
  • nany integer (n ∈ ℤ) — the number of whole turns

Written out: x = arcsin a + 2πn and x = π − arcsin a + 2πn. For |a| > 1 there are no roots.

Equations sin x = a

Solve:
1) sin x = √3/2
2) sin x = −1/2
3) sin 2x = √2/2
4) sin x = 0.3

Show solution
1) arcsin(√3/2) = π/3: x = (−1)ⁿ · π/3 + πn, that is π/3 + 2πn and 2π/3 + 2πn.
2) arcsin(−1/2) = −π/6: x = (−1)ⁿ · (−π/6) + πn = (−1)ⁿ⁺¹ · π/6 + πn. As two series: −π/6 + 2πn and 7π/6 + 2πn.
3) 2x = (−1)ⁿ · π/4 + πn. Divide both sides by 2 — πn too: x = (−1)ⁿ · π/8 + πn/2.
4) x = (−1)ⁿ · arcsin 0.3 + πn (arcsin 0.3 ≈ 0.305).
Interactive
Loading simulation…
Move the slider a: while |a| < 1 the line y = a cuts the sine wave at two points in every period (two series of roots), at a = ±1 it just touches it (one series), and for |a| > 1 there is no intersection — and no roots.

On each branch (−π/2, π/2) the tangent takes every real value exactly once, and its period is π. So for any a the equation tan x = a has a single series of roots: the root on the branch is arctan a, and the others repeat with step π. The same goes for the cotangent.

tan x = a: x = arctan a + πn; cot x = a: x = arccot a + πn, n ∈ ℤ
where:
  • aany real number
  • nany integer (n ∈ ℤ) — the number of whole turns

There is no condition on a, because the tangent and the cotangent take every real value; the step of the series is π, not 2π.

Equations tan x = a and cot x = a

Solve:
1) tan x = √3
2) tan(2x + π/4) = −1
3) cot x = 1
4) tan x = −2

Show solution
1) arctan √3 = π/3: x = π/3 + πn.
2) 2x + π/4 = arctan(−1) + πn = −π/4 + πn ⇒ 2x = −π/2 + πn ⇒ x = −π/4 + πn/2.
3) arccot 1 = π/4: x = π/4 + πn.
4) arctan(−2) = −arctan 2: x = −arctan 2 + πn (arctan 2 ≈ 1.107).

For a = 0 and a = ±1 the general formulas still give the right answer, but the roots can be written more briefly. We already know these points from the graphs: the zeros, peaks and troughs of the sine and cosine. They are worth knowing by heart:

asin x = acos x = a
0x = πnx = π/2 + πn
1x = π/2 + 2πnx = 2πn
−1x = −π/2 + 2πnx = π + 2πn
Special cases (n ∈ ℤ). Note: for sin x = 0 and cos x = 0 the step is π, for ±1 it is 2π.
Special cases

Solve:
1) sin 3x = 0
2) cos(x + π/4) = 1
3) sin(x/2) = −1

Show solution
1) 3x = πn ⇒ x = πn/3.
2) x + π/4 = 2πn ⇒ x = −π/4 + 2πn.
3) x/2 = −π/2 + 2πn ⇒ x = −π + 4πn. The general formula would give the same result, only by a longer route.

Equations that reduce to the simplest ones

A harder trigonometric equation is reduced to one or several of the simplest ones by four main methods. The general algorithm:

  1. 1
    Bring to one function of one argument

    Use the identities: sin²x = 1 − cos²x, cos 2x = 1 − 2sin²x = 2cos²x − 1, sin 2x = 2 sin x cos x and so on.

  2. 2
    Choose a method

    A function and its square — substitution; a common factor — factoring; all terms of the same degree — a homogeneous equation; a sin x + b cos x or sin²x — identities.

  3. 3
    Solve the simplest equations

    After substituting t = sin x or t = cos x, check |t| ≤ 1 and drop equations such as sin x = 2.

  4. 4
    Check the domain

    If there is a denominator, tan or cot, remove the roots that make them zero or meaningless.

  5. 5
    Write the answer

    Write all the series with n ∈ ℤ; if required, select the roots in the given interval.

Reducing to a quadratic. If the equation contains a function and its square, the substitution t = sin x (or t = cos x, t = tan x) turns it into a quadratic equation. For sin x and cos x we keep only the roots with −1 ≤ t ≤ 1.

Reducing to a quadratic

Solve:
1) 2sin²x − sin x − 1 = 0
2) 2cos²x + 3 sin x − 3 = 0
3) cos 2x + 5 cos x + 3 = 0

Show solution
1) t = sin x: 2t² − t − 1 = 0, t = (1 ± 3)/4, so t = 1 or t = −1/2.
sin x = 1 ⇒ x = π/2 + 2πn; sin x = −1/2 ⇒ x = (−1)ⁿ⁺¹ · π/6 + πn.
2) cos²x = 1 − sin²x: 2 − 2sin²x + 3 sin x − 3 = 0 ⇒ 2sin²x − 3 sin x + 1 = 0. t = sin x: t = 1 or t = 1/2.
x = π/2 + 2πn; x = (−1)ⁿ · π/6 + πn.
3) cos 2x = 2cos²x − 1: 2cos²x + 5 cos x + 2 = 0. t = cos x: t = −1/2 or t = −2.
t = −2 does not fit: a cosine cannot be less than −1. cos x = −1/2 ⇒ x = ±2π/3 + 2πn.

Factoring. We bring the equation to the form f · g = 0: if a product is zero, one of the factors is zero. A common factor is taken out of the bracket, and a sum is turned into a product. Never divide both sides by sin x or cos x — the roots that make that factor zero would be lost.

Factoring

Solve:
1) sin 2x = cos x
2) sin x + sin 3x = 0
3) sin²x = sin x · cos x

Show solution
1) 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.
Dividing both sides by cos x would have lost the first series.
2) Turn the sum into a product: 2 sin 2x · cos x = 0.
sin 2x = 0 ⇒ x = πn/2; cos x = 0 ⇒ x = π/2 + πn. The second series is inside the first, so the answer is x = πn/2.
3) sin x · (sin x − cos x) = 0.
sin x = 0 ⇒ x = πn; sin x = cos x ⇒ tan x = 1 ⇒ x = π/4 + πn. Here dividing by cos x is allowed: if cos x = 0, then sin x = cos x would give sin x = 0 as well, which is impossible.
Definition
Homogeneous equation

An equation in which all terms have the same degree in sin x and cos x: first degree a sin x + b cos x = 0, second degree a sin²x + b sin x cos x + c cos²x = 0. Dividing it by cos x (the second-degree one by cos²x) gives an equation in the tangent.

Why is this division allowed? If a ≠ 0, the points where cos x = 0 cannot be roots: there sin x = ±1, and the left side becomes a sin x (or a sin²x), which is not zero. So no roots are lost by dividing.

Homogeneous equations

Solve:
1) sin x − √3 cos x = 0
2) sin²x − 4 sin x cos x + 3cos²x = 0
3) 3sin²x + sin x cos x = 2

Show solution
1) Divide by cos x: tan x − √3 = 0 ⇒ x = π/3 + πn.
2) Divide by cos²x: tan²x − 4 tan x + 3 = 0 ⇒ tan x = 1 or tan x = 3.
x = π/4 + πn; x = arctan 3 + πn.
3) The equation is not homogeneous: the 2 on the right has degree zero. We “make it homogeneous”: 2 = 2(sin²x + cos²x).
3sin²x + sin x cos x − 2sin²x − 2cos²x = 0 ⇒ sin²x + sin x cos x − 2cos²x = 0.
Divide by cos²x: tan²x + tan x − 2 = 0 ⇒ tan x = 1 or tan x = −2.
x = π/4 + πn; x = −arctan 2 + πn.

Using identities. The equation a sin x + b cos x = c is reduced to a single sine by the auxiliary angle method, and in equations with sin²x or cos²x the power-reduction formula helps — both are in the lesson “Addition, double-angle and half-angle formulas”.

Auxiliary angle and power reduction

Solve:
1) sin x + √3 cos x = 1
2) 4sin²x = 1

Show solution
1) R = √(1 + 3) = 2: sin x + √3 cos x = 2 sin(x + π/3). So sin(x + π/3) = 1/2.
x + π/3 = π/6 + 2πn ⇒ x = −π/6 + 2πn; x + π/3 = 5π/6 + 2πn ⇒ x = π/2 + 2πn.
Check: at x = π/2 we get 1 + √3 · 0 = 1.
2) sin²x = (1 − cos 2x)/2: 2(1 − cos 2x) = 1 ⇒ cos 2x = 1/2 ⇒ 2x = ±π/3 + 2πn ⇒ x = ±π/6 + πn.
This is shorter than solving sin x = 1/2 and sin x = −1/2 separately: one formula gives all the roots in the four quadrants.

Selecting roots in an interval

The answer to an equation is an infinite set, but a problem often asks only for the roots in a certain segment. There are three ways to find them: substitute n = …, −1, 0, 1, 2, … one after another; put the series into a double inequality and find the integer values of n; or mark the roots on the unit circle or on the graph. In equations with a denominator, tan or cot, the roots must also be checked against the domain.

By substituting n

Find the roots of 2cos²x − cos x − 1 = 0 in the segment [−π, π].

Show solution
t = cos x: 2t² − t − 1 = 0 ⇒ t = 1 or t = −1/2.
cos x = 1 ⇒ x = 2πn. For n = 0 we get x = 0, inside the segment; for n = ±1 we get ±2π, outside it.
cos x = −1/2 ⇒ x = ±2π/3 + 2πn. For n = 0, ±2π/3 lie in the segment; for n = ±1 the values leave it.
Answer: −2π/3, 0, 2π/3.
With a double inequality

How many roots does tan 2x = 1 have in the segment [0, 2π]? Find their sum.

Show solution
2x = π/4 + πn ⇒ x = π/8 + πn/2.
0 ≤ π/8 + πn/2 ≤ 2π. Multiply every part by 8/π: 0 ≤ 1 + 4n ≤ 16 ⇒ −1/4 ≤ n ≤ 15/4.
Integers n: 0, 1, 2, 3 — 4 roots: π/8, 5π/8, 9π/8, 13π/8.
Sum: (1 + 5 + 9 + 13)π/8 = 28π/8 = 7π/2.
Checking against the domain

Solve sin 2x / (1 + cos x) = 0.

Show solution
A fraction is zero when its numerator is zero and its denominator is not.
sin 2x = 0 ⇒ x = πn/2: the points 0, π/2, π, 3π/2 on the unit circle.
1 + cos x ≠ 0 ⇒ cos x ≠ −1 ⇒ x ≠ π + 2πn: the point π is removed.
The points 0, π/2 and 3π/2 remain. Answer: x = 2πn; x = π/2 + πn.

The simplest trigonometric inequalities

In an inequality we look not for single points but for a whole arc. sin x > a means that the point on the unit circle lies above the line y = a; cos x > a means that it lies to the right of the line x = a. We find the arc, write down its ends and add whole turns.

  1. 1
    Draw the line

    For sin draw the horizontal line y = a, for cos the vertical line x = a.

  2. 2
    Find the intersection points

    For sin: arcsin a and π − arcsin a; for cos: arccos a and −arccos a.

  3. 3
    Choose the arc

    Mark the arc that satisfies the inequality: for “>” the arc above (to the right of) the line, for “<” the arc below (to the left of) it.

  4. 4
    Go round the arc anticlockwise

    Go from the start of the arc to its end anticlockwise: the left end must be smaller than the right end. If it is not, subtract 2π from the left end.

  5. 5
    Add the period

    Add 2πn to both ends (πn for tan); for “≥” and “≤” the ends are included.

sin x > a: arcsin a + 2πn < x < π − arcsin a + 2πncos x > a: −arccos a + 2πn < x < arccos a + 2πntan x > a: arctan a + πn < x < π/2 + πnsin x > a: arcsin a + 2πn < x < π − arcsin a + 2πncos x > a: −arccos a + 2πn < x < arccos a + 2πntan x > a: arctan a + πn < x < π/2 + πn
where:
  • a|a| < 1 for sin and cos, any number for tan
  • nany integer (n ∈ ℤ) — the number of whole turns

For inequalities with “<” take the rest of the circle. For |a| ≥ 1 the answer can be read off the graph at a glance: for example, sin x > 1 has no solutions, while sin x ≤ 1 holds for every x.

Interactive
Loading simulation…
Drag the point and watch sin α: while α is between 30° and 150°, sin α ≥ 1/2 — this is the arc of the inequality sin x ≥ 1/2. Find the arc of cos α < √2/2 as well: from 45° to 315°.
Inequalities on the unit circle

Solve:
1) sin x ≥ 1/2
2) cos x < √2/2
3) sin x < −1/2
4) tan x ≤ 1

Show solution
1) The line y = 1/2 meets the circle at π/6 and 5π/6; the arc above it runs from π/6 to 5π/6: π/6 + 2πn ≤ x ≤ 5π/6 + 2πn.
2) The arc to the left of the line x = √2/2 starts at π/4 and goes anticlockwise to 7π/4: π/4 + 2πn < x < 7π/4 + 2πn.
3) The arc below the line y = −1/2 runs from 7π/6 to 11π/6: 7π/6 + 2πn < x < 11π/6 + 2πn (another way to write it: −5π/6 + 2πn < x < −π/6 + 2πn).
4) The tangent increases on the branch (−π/2, π/2) and equals 1 at π/4: −π/2 + πn < x ≤ π/4 + πn. The left end is not included — the tangent is undefined there.
Inequalities with a composite argument, and an application

1) Solve 2 cos 2x > 1.
2) Solve sin x > 2; sin x ≤ 1; cos x > −1.
3) With h(t) = 30 − 28 cos(πt/15), how long is Elvin’s cabin higher than 44 m during each turn?

Show solution
1) cos 2x > 1/2 ⇒ −π/3 + 2πn < 2x < π/3 + 2πn. Divide by 2: −π/6 + πn < x < π/6 + πn.
2) sin x > 2 — no solutions; sin x ≤ 1 — every x; cos x > −1 — only the points where cos x = −1 drop out: x ≠ π + 2πn.
3) 30 − 28 cos(πt/15) > 44 ⇒ cos(πt/15) < −1/2 ⇒ 2π/3 + 2πn < πt/15 < 4π/3 + 2πn.
Multiply by 15/π: 10 + 30n < t < 20 + 30n. In every 30-minute turn the cabin is above 44 m for 10 minutes — from minute 10 to minute 20.

Key points

  • arcsin a ∈ [−π/2, π/2], arccos a ∈ [0, π], arctan a ∈ (−π/2, π/2); arcsin(−a) = −arcsin a, arccos(−a) = π − arccos a.
  • sin x = a (|a| ≤ 1): x = (−1)ⁿ arcsin a + πn; cos x = a: x = ±arccos a + 2πn; tan x = a: x = arctan a + πn; for |a| > 1, sin x = a and cos x = a have no roots.
  • Special cases: sin x = 0 ⇔ x = πn; cos x = 0 ⇔ x = π/2 + πn; sin x = ±1 ⇔ x = ±π/2 + 2πn; cos x = 1 ⇔ x = 2πn; cos x = −1 ⇔ x = π + 2πn.
  • Four methods: reduce to a quadratic (check |t| ≤ 1), factor (never divide by sin x or cos x!), divide a homogeneous equation by cosⁿx, and use the auxiliary angle or power reduction.
  • Roots are selected by substituting n or with a double inequality; in equations with a denominator the domain is checked.
  • Inequalities are solved on the unit circle: a horizontal line for sin, a vertical one for cos; the arc is written anticlockwise and 2πn is added.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
What is arccos(−1/2)?