Trigonometric equations and inequalities

The app brings an equation to basic equations such as sin x = a and solves them on the unit circle: it finds the angles of one turn and adds multiples of the period (x = π/6 + 2kπ, k ∈ ℤ). The circle is drawn right in the steps.

It prefers the known formulas – double and half angle, sin²x + cos²x = 1, sum to product –, then factoring, the substitution t = sin x, dividing a homogeneous equation by cosⁿx and the auxiliary angle for a·sin x + b·cos x = c. The result is also checked numerically over a whole period.

An unknown written as a Latin letter (x, t) is in radians, a Greek one (alpha, β, …) in degrees – as is an input with the ° sign (sin(x + 30°)). To get the roots in a given interval, add the interval after a comma or on the next line.

An inequality is solved by an arc of the unit circle, a more complex one by the zeros on one period; the result is intervals with the period.

Examples

Two branches of solutions, the angles on the unit circle.
Input: sin x = 1/2
sin x = 1/2
sin x = 1/2sin x = 1/2
The substitution t = sin x and a quadratic equation.
Input: 2sin^2 x - sin x - 1 = 0
2sin^2 x - sin x - 1 = 0
2sin^2 x - sin x - 1 = 02sin^2 x - sin x - 1 = 02sin^2 x - sin x - 1 = 0
The formula for sin 2x and factoring.
Input: sin 2x = cos x
sin 2x = cos x
sin 2x = cos xsin 2x = cos xsin 2x = cos xsin 2x = cos x
The auxiliary angle: sin x + cos x = √2·sin(x + π/4).
Input: sin x + cos x = 1
sin x + cos x = 1
sin x + cos x = 1sin x + cos x = 1sin x + cos x = 1
A Greek letter – the result in degrees.
Input: sin alpha + cos alpha = sqrt(2)
sin alpha + cos alpha = sqrt(2)
sin alpha + cos alpha = sqrt(2)sin alpha + cos alpha = sqrt(2)
Only the roots in the given interval.
Input: sin x = 1/2, x ∈ <0; 4pi)
sin x = 1/2, x ∈ <0; 4pi)
sin x = 1/2, x ∈ <0; 4pi)sin x = 1/2, x ∈ <0; 4pi)sin x = 1/2, x ∈ <0; 4pi)
An arc of the unit circle, an interval around zero.
Input: cos x > 1/2
cos x > 1/2
cos x > 1/2cos x > 1/2
The zeros on a period and the sign between them.
Input: 2sin^2 x - sin x - 1 > 0
2sin^2 x - sin x - 1 > 0
2sin^2 x - sin x - 1 > 02sin^2 x - sin x - 1 > 02sin^2 x - sin x - 1 > 0

Known limitations