From One Cycle to Any Interval
On one cycle, a basic trigonometric equation has only a small number of solutions. But an interval in a problem may begin below zero, extend beyond a full cycle, or include an endpoint that is itself a solution. Periodicity lets us find all solutions: first describe the repeated solution angles, then keep exactly those that lie in the stated interval.
The interval check is essential. A list of angles that solve the equation is not yet the answer if some lie outside the interval, and an angle at an excluded endpoint must not be included.
Periodic Solution Families
The sine and cosine functions repeat after \(2\pi\), while the tangent function repeats after \(\pi\). Thus, adding an integer multiple of the appropriate period to a solution produces another solution. Here \(k\) denotes any integer.
For \(\sin x=c\), a real solution exists only when \(-1\le c\le 1\). Let \(\alpha=\arcsin(c)\), where the principal inverse sine has values in \([-\pi/2,\pi/2]\). The solution families are \(x=\alpha+2\pi k\) and \(x=\pi-\alpha+2\pi k\). At \(c=1\) or \(c=-1\), these two families describe the same angles, so repeated answers must be listed only once.
For \(\cos x=c\), a real solution exists only when \(-1\le c\le 1\). Let \(\alpha=\arccos(c)\), where the principal inverse cosine has values in \([0,\pi]\). The families are \(x=\alpha+2\pi k\) and \(x=-\alpha+2\pi k\), again removing duplicates when the two families coincide. For \(\tan x=c\), let \(\alpha=\arctan(c)\), whose principal values lie in \((-\pi/2,\pi/2)\). Its family is \(x=\alpha+\pi k\).
Why the Families Give Every Solution
Periodicity explains why the listed angles solve the equation, but we also need to know that the families do not miss any solutions. The reason is that any angle can be reduced to one cycle by subtracting an integer number of periods. The one-cycle solutions then provide representatives for the repeated families.
Use \(T=2\pi\) for sine and cosine, and \(T=\pi\) for tangent.
For any real \(x\), there is an integer \(k\) such that \(r=x-kT\) belongs to \([0,T)\). Then \(x=r+kT\).
Periodicity gives \(f(x)=f(r)\), for the relevant trigonometric function \(f\). Therefore, if \(x\) solves the equation, \(r\) must be one of its solution representatives in \([0,T)\). Conversely, shifting any such representative by \(kT\) preserves the function value.
This also highlights a useful detail: the representative \(r\) used in \([0,T)\) must actually lie in that interval. A listed angle outside the interval must first be shifted by a whole period to obtain its one-cycle representative. The families above describe all shifts, while the interval check selects the requested ones.
Filtering a Family Against an Interval
For a family \(x=r+kT\) and a closed interval \([a,b]\), require \(a\le r+kT\le b\). Rearranging gives bounds on the integer \(k\):
Only the integers in those bounds are used. For an open endpoint, replace the corresponding weak inequality by a strict one. For example, on \((a,b]\), require \(a<r+kT\le b\). With more than one solution family, do this for each family, combine the retained angles, and remove duplicates.
It is often simplest to test the integer values near the bounds rather than round the bounds carelessly. The integer \(k\) itself must be an integer; an ordinary decimal value of \(k\) is not a solution shift.
Worked Examples
Worked Example: Sine Solutions on an Interval Crossing Zero
Find all solutions of \(\sin x=\frac12\) on \([-\pi,5\pi/2)\).
The one-cycle angles are \(\pi/6\) and \(5\pi/6\), since \(\sin(\pi/6)=\sin(5\pi/6)=1/2\). The period is \(2\pi\), so the families are
For the first family, apply the interval bounds:
The integers in these bounds are \(k=0,1\). They give \(x=\pi/6\) and \(x=13\pi/6\). For the second family:
The only integer is \(k=0\), giving \(x=5\pi/6\). The complete answer is \(\pi/6,5\pi/6,13\pi/6\). Each is in the interval, and their sine values are all \(1/2\). The next shift of either family falls outside the interval.
Worked Example: Cosine Solutions on a Closed Interval
Find all solutions of \(\cos x=-\frac{\sqrt{2}}{2}\) on \([\pi/4,9\pi/4]\).
The one-cycle solutions are \(3\pi/4\) and \(5\pi/4\), and the period is \(2\pi\). Thus the families are \(x=3\pi/4+2\pi k\) and \(x=5\pi/4+2\pi k\). For the first family, the interval condition becomes
Only \(k=0\) is an integer in this range, so this family contributes \(3\pi/4\). For the second family,
Again, only \(k=0\) is allowed, giving \(5\pi/4\). Both endpoints of the interval are included, but neither is a solution. Checking the two answers gives \(\cos(3\pi/4)=\cos(5\pi/4)=-\sqrt{2}/2\). No other integer shifts lie in the interval.
Worked Example: Tangent Solutions on an Interval with an Open Endpoint
Find all solutions of \(\tan x=\sqrt{3}\) on \((-\pi,3\pi/2]\).
The principal angle is \(\pi/3\), since \(\tan(\pi/3)=\sqrt{3}\). Tangent has period \(\pi\), so \(x=\pi/3+\pi k\). Apply the interval bounds:
The integers satisfying this are \(k=-1,0,1\). Substitution into the family gives \(x=-2\pi/3,\pi/3,4\pi/3\). All three lie strictly above \(-\pi\) and at or below \(3\pi/2\). Their tangent values are \(\sqrt{3}\), because each differs from \(\pi/3\) by an integer multiple of \(\pi\). The next shifts, for \(k=-2\) and \(k=2\), fall outside the interval. Thus these are all the solutions.
Worked Example: An Endpoint That Is a Solution
Find all solutions of \(\sin x=0\) on \([0,2\pi)\).
The sine solutions are \(x=k\pi\) for integers \(k\). Requiring \(0\le k\pi<2\pi\) gives \(0\le k<2\), so \(k=0,1\). The solutions are \(0\) and \(\pi\). Both satisfy \(\sin x=0\), and both belong to the interval. Although \(\sin(2\pi)=0\), \(2\pi\) is not included because the right endpoint is open.
Representatives and a Common Pitfall
A common mistake is to use a negative angle as if it were already the one-cycle representative in \([0,2\pi)\). For instance, \(-\pi/3\) is a valid solution angle for \(\cos x=1/2\), but it is not in \([0,2\pi)\). Its representative there is found by adding one period:
Similarly, \(-\pi/2\) is not in \([0,2\pi)\); adding \(2\pi\) gives \(3\pi/2\), the one-cycle solution for \(\sin x=-1\). Treating either negative angle itself as a representative in \([0,2\pi)\) can cause a solution to be missed or counted in the wrong cycle.
A Reliable Checklist
For an equation on a specified interval, use the following sequence. It separates finding the repeated solutions from deciding which of them are allowed.
Use the one-cycle solutions for the given sine, cosine, or tangent equation.
Use period \(2\pi\) for sine and cosine and period \(\pi\) for tangent.
Solve the resulting inequalities for integer \(k\), using strict inequalities at open endpoints.
Substitute each retained angle into the equation, confirm it belongs to the interval, and remove any duplicates.
The method works because each family contains exactly the period-shifts of a base solution, and the interval inequalities select precisely the shifts that lie in the requested region. When an interval is unbounded, the same families and membership test apply; an unbounded interval may contain infinitely many solutions.
Check Your Understanding
For each question, find every solution in the stated interval and check the endpoint conventions.
- Solve \(\cos x=0\) on \([0,2\pi)\).
- Solve \(\sin x=-1\) on \([-2\pi,0]\).
- Solve \(\tan x=1\) on \([0,2\pi]\).
- Why must a negative starting angle be shifted before it can be called a representative in \([0,2\pi)\)?
- What changes in the interval inequality when the right endpoint is open rather than closed?
