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Trigonometric equations in radians

Solve sine, cosine and tangent equations in radians. Find every solution in positive or negative intervals, handle endpoints and check the sine-rule ambiguous case.

Before you startExact radian values, trigonometric graphs and inverse trigonometric functions

01 / An equation asks for every matching input

One calculator angle is usually only the start.

Solving sin x =1/2 means finding every angle whose sine is 1/2 in the stated interval. A graph shows two intersections in one full turn: x =π/6 and x =5π/6.

Identify the function and value → find a reference or principal angle → generate all branches → filter the interval.

Use the controls to change the equation or interval. Written methods and worked solutions remain available independently of the optional animation.

Count every intersectionExplore

2 solutions: π/6, 5π/6.

Interval: 0 ≤ x ≤ 2π. Both endpoints are included.

Each highlighted point is one solution in the selected interval. Dashed vertical lines mark tangent’s excluded inputs. Open endpoint circles are not solutions.

FunctionRight-hand valueInterval

Watch the second sine solution appear

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

02 / Use inverse trig with the right angle mode

The inverse function returns a principal value.

Set a calculator to radians before using inverse trig for a radian answer. Principal values are arcsin k in [−π/2,π/2], arccos k in [0,π], and arctan k in (−π/2,π/2). These ranges explain why an inverse button cannot list every solution automatically.

Solve sin x =0.3 for 0≤x≤2πWorked example

α =arcsin(0.3) ≈0.304693

Keep this unrounded value in later calculations.

x =α or π−α

Sine is positive in quadrants I and II.

x ≈0.304693 or 2.836900

State radians; round only the final answers.

If sin x =1.2 or cos x =−1.4, there is no real solution: sine and cosine lie in [−1,1]. Tangent does not have that restriction.

03 / Sine uses equal heights on the circle

The two branches are α and π−α, repeated every 2π.

sin x =k: x =α+2nπ or x =π−α+2nπ
α =arcsin k, n an integer, |k|≤1

Solve sin x =−√3/2 on 0≤x<2πWorked example

Reference angle =π/3

The sine is negative in quadrants III and IV.

x =π+π/3 or 2π−π/3

Both lie in the requested interval.

x =4π/3 or 5π/3

Check both sines are −√3/2.

At k=1 or k=−1, the two branches coincide after a full-turn shift. Do not count the same angle twice.

04 / Cosine uses equal horizontal coordinates

Reflect the principal angle through a full turn.

cos x =k: x =±β+2nπ
β =arccos k, n an integer, |k|≤1

Solve cos x =−1/2 on 0≤x≤2πWorked example

β =2π/3

This principal value is in quadrant II.

Other angle =2π−2π/3 =4π/3

Cosine is also negative in quadrant III.

x =2π/3 or 4π/3

Neither endpoint is a solution.

The identity cos(π/2−x)=sin x can also convert a phase-related expression, but changing the function does not remove the need to check the interval.

05 / Tangent repeats after π

Generate solutions in steps of a half-turn.

tan x =k: x =γ+nπ
γ =arctan k, n an integer

Solve tan x =−1 on 0≤x<2πWorked example

γ =−π/4

The principal value is negative.

Add π to reach 3π/4, then add π again

This gives the next solution at 7π/4.

x =3π/4 or 7π/4

Tangent is undefined at π/2 and 3π/2; those are not candidates.

A tangent curve crossing a vertical asymptote is not an intersection with the equation line. An asymptote has no function value.

06 / Negative intervals need negative angles

Use periodic shifts rather than assuming one positive turn.

Solve sin x =1/2 for −2π≤x<2πWorked example

One-turn roots: π/6 and 5π/6

Subtract 2π from each to reach the previous turn.

Previous-turn roots: −11π/6 and −7π/6

Both remain at or above −2π.

Complete set: −11π/6, −7π/6, π/6, 5π/6

Adding or subtracting another full turn would leave the interval.

Solve cos x =−1/2 on −π≤x≤πWorked example

Use x =±2π/3+2nπ

The n=0 pair already lies in the interval.

x =−2π/3 or 2π/3

The answer need not contain only positive angles.

07 / Count interval endpoints explicitly

The same point on a circle can represent different allowed inputs.

Solve sin x =0 on 0≤x≤2πWorked example

General roots are x=nπ

List all allowed integers n.

n=0,1,2

Both endpoints are included.

x=0, π, 2π

Zero and 2π are distinct inputs even though they share a circle position.

On 0≤x<2π, remove 2π. On −2π≤x<2π, the sine-zero set is −2π, −π, 0, π. Open endpoints are excluded even when the graph would meet the target there.

For cos x=1 on 0≤x≤2π, both 0 and 2π count. For sin x=1, the repeated algebraic branches give only π/2 in that interval.

08 / Describe all angles, then select integers

General solutions organise a long interval.

Solve tan x =√3 for −π≤x<3πWorked example

x =π/3+nπ

The principal value is π/3.

−π≤π/3+nπ<3π

Subtract π/3 and divide by positive π.

−4/3≤n<8/3, so n=−1,0,1,2

Only integer n are allowed.

x=−2π/3, π/3, 4π/3, 7π/3

Substitute into the original interval as a final check.

General formulas can duplicate roots for extreme sine/cosine values. A final answer is a set of distinct angles, not a count of algebraic branches.

09 / A sine-rule angle must still fit a triangle

Two sine values do not always mean two possible triangles.

Use a and b for side lengths opposite angles A and B. The sine rule gives sin B =b sin A/a. A candidate B must satisfy 0<B<π and A+B<π, so the remaining angle is positive.

a=8, b=10 and A=π/6Worked example

sin B =10(1/2)/8 =5/8

Let β=arcsin(5/8).

B=β or π−β

Here β≈0.675132 is greater than A≈0.523599.

Both give A+B<π

The remaining angles are 5π/6−β and β−π/6, both positive.

a=10, b=6 and A=π/3Worked example

sin B =3√3/10

Let β=arcsin(3√3/10).

β<π/3

Solve cos(π/2−x)=1/2 for 0≤x<2π.

Hint

Use cos(π/2−x)=sin x.

Worked solution

sin x=1/2, so x=π/6, 5π/6.

11 / Recap

A complete answer respects both periodicity and the interval.

Use radian mode and retain unrounded principal values.
Sine and cosine repeat every 2π; tangent every π.
Generate both sine/cosine branches where distinct.
Include negative angles when required.
Check open and closed endpoints and remove duplicate roots.
Sine-rule candidates must leave a positive third angle.

Section 1 of 11 · An equation asks for every matching input