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Solving trigonometric equations

Solve sine, cosine and tangent equations in a given interval. Find all solutions, understand inverse-calculator ranges and check endpoint and domain restrictions.

Before you startUnit-circle signs, exact values and basic trigonometric graphs

01 / Read the interval

An equation needs a range of angles.

sin x = 1/2 has infinitely many solutions. A question such as “solve for 0° ≤ x < 360°” asks you to keep only the angles inside that interval. Its left endpoint is included and its right endpoint is excluded.

Rearrange to sin x = k, cos x = k or tan x = k first. Sine and cosine take values from −1 to 1, so a target outside that range gives no real solutions. Tangent can equal any real number, but it is undefined at 90° + 180°n.

Find every intersection in the intervalChoose and compare

Equation and interval

sin x = 0.5, x in [0°, 360°). Solutions, rounded to four decimal places: 30°, 150°.

Watch all the sine intersections appear

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / What the calculator returns

Inverse trig chooses one principal value.

sin⁻¹ k returns an angle in [−90°, 90°]
cos⁻¹ k returns an angle in [0°, 180°]
tan⁻¹ k returns an angle in (−90°, 90°)

The sine and cosine inverse inputs must lie in [−1,1]. Tangent’s inverse range excludes both endpoints.

Use degree mode. sin⁻¹ means inverse sine (arcsin), not 1/sin. For example, sin⁻¹(1/2) = 30°, but 150° has the same sine. tan⁻¹(−1) = −45°, even if the requested interval starts at 0°.

Keep the unrounded principal value while generating other candidates, then round the final answers to the requested accuracy.

03 / Sine equations

A horizontal line usually meets two points per turn.

For sin x = k, let α = sin⁻¹ k. The complete families are:

x = α + 360°n
or x = 180° − α + 360°n, where n is an integer

For sin x = −1/2, α = −30°. The two families are −30° + 360°n and 210° + 360°n. In −360° ≤ x ≤ 360°, these give:

x = −150°, −30°, 210°, 330°

For k = 1 or −1 the two families overlap, so count repeated values only once. For k = 0, they give the different multiples 0°, 180°, 360° and so on.

04 / Cosine equations

Reflect the principal angle in the horizontal axis.

For cos x = k, let α = cos⁻¹ k. The complete families are:

x = α + 360°n
or x = −α + 360°n, where n is an integer

For cos x = −√2/2, α = 135°. In −180° ≤ x ≤ 540°:

x = −135°, 135°, 225°, 495°

At k = 1 or −1, remove duplicate angles from the two families. The signs ± belong to the angle, not to the given cosine value.

05 / Tangent equations

One family repeats every 180°.

For tan x = k, let α = tan⁻¹ k. Then:

x = α + 180°n, where n is an integer

For tan x = −√3, α = −60°. In −270° ≤ x ≤ 270° the answers are:

x = −240°, −60°, 120°

Do not use the sine supplement rule 180° − α for tangent. Tangent has the same sign in opposite quadrants and repeats after half a turn.

06 / Endpoints and counting

The bracket decides whether a boundary solution stays.

For sin x = 0 on 0° ≤ x ≤ 360°, the answers are 0°, 180° and 360°. On 0° ≤ x < 360°, keep only 0° and 180°. Although 0° and 360° reach the same point on the circle, they are different numbers in a closed interval.

Before calculating, a graph can check the expected count. For −1 < k < 1, a sine or cosine curve has two intersections per full period on a half-open interval of length 360°. At an extreme k = ±1 it has one. Tangent has one solution per half-open period of length 180°.

Prove there are no solutions

5sin x = 7 is impossible because sin x ≤ 1. Also 2sin x + 3cos x + 6 = 0 is impossible: each term is at least −2 and −3 respectively, so the left side is at least 1. This bound is sufficient here; it does not claim the two minima occur together.

07 / Sine equals a multiple of cosine

Check zero cosine before dividing.

Solve 2sin x = √3 cos x for 0° ≤ x < 360°. If cos x = 0, the original equation would require sin x = 0 too, which cannot happen. So division by cos x is safe for every possible solution.

tan x = √3/2
α = tan⁻¹(√3/2) ≈ 40.8934°
x ≈ 40.9°, 220.9°

A useful check is substitution into the original equation with unrounded values. The reciprocal error tan x = 2/√3 would give different angles.

sin x = 2cos x, 0° ≤ x < 360°

Hint

Check cos x = 0, then divide.

Worked solution

tan x = 2
x ≈ 63.4°, 243.4°

At cos x = 0 the original equation fails, so no solutions were lost.

09 · No real tangent

tan x + 1/tan x = 0

Hint

The original requires tangent to exist and be non-zero.

Worked solution

tan² x + 1 = 0

No real solutions, because a real square cannot equal −1.

09 / Recap

A principal value starts the search.

Rearrange and check whether the ratio is possible.
Use the correct inverse range and degree mode.
Write both sine/cosine families or the single tangent family.
Generate all candidates across the given interval.
Check open and closed endpoints, remove duplicates, and round last.

Next: multiple-angle and shifted-angle equations →

Section 1 of 9 · Read the interval