Solving trigonometric equations
An inverse key gives back one angle. When a sine or cosine equals some number, how do you find every angle that works?
Where you are 4 inside
What this area is about
A trig equation runs evaluation backward: the value is given and the angles are wanted. Two facts shape every answer. The functions are periodic (sine and cosine every 2\pi, tangent every \pi, in radians), so one solution brings infinitely many. And an inverse function returns only the one angle in its restricted range, so the circle’s symmetry has to supply the rest. An answer is either every solution, written as families with k any integer, or the finite list inside a stated interval such as [0, 2\pi).
How these fit together
The core procedure. Solving basic trig equations gets \sin x, \cos x or \tan x alone, finds the solutions in one turn, then extends them by the period. Everything else reduces to it.
Harder equations. Equations with a multiple angle such as \sin(3x) = c solve for the inner angle, stretch the interval to match, and divide at the end. Factoring, quadratics and identities breaks an equation with two trig terms, a square or a double angle into basic ones; squaring both sides can add false solutions, so after squaring every candidate is checked.
Beyond equality. Trigonometric inequalities find whole stretches of angles where a function stays above or below a level, cut at the crossings of the matching equation and at any point where the function is undefined.
A path through it
Basic equations first: the multiple-angle lesson builds on them, and the factoring and inequality lessons refer back to them. The other three follow; none of them is a prerequisite of another. The factoring lesson also needs the double angle formulas.