Trigonometric Equations

Suppose someone asks: find an angle x such that \sin x = \frac{1}{2}.

You know the answer immediately: x = 30°, or in radians, x = \frac{\pi}{6}. That is the angle you memorised from the standard table.

But here is the thing. Try x = 150° (which is \frac{5\pi}{6}). You get \sin 150° = \frac{1}{2}. That also works. Try x = 390° (which is \frac{\pi}{6} + 2\pi). That works too. Try x = -\frac{11\pi}{6}. Also works.

In fact, there are infinitely many angles that satisfy \sin x = \frac{1}{2}. Not two, not ten, but infinitely many. The sine function keeps repeating with period 2\pi, and every time it completes a cycle, the same values come back.

This is what makes trigonometric equations different from the algebraic equations you have solved before. A quadratic equation has two solutions. A linear equation has one. But \sin x = \frac{1}{2} has an infinite family of solutions, and you need a single compact formula that describes all of them at once.

That formula is called the general solution, and learning to write it down correctly is the entire point of this article.

Why trig equations have infinitely many solutions

The reason is periodicity. Plot y = \sin x on a graph. The curve oscillates between -1 and +1, completing one full cycle every 2\pi radians. Now draw a horizontal line at height y = \frac{1}{2}. That line cuts through the sine curve over and over again — twice per cycle, once on the way up, once on the way down.

Every intersection point on that graph is a solution of \sin x = \frac{1}{2}. The intersections come in pairs — two per period — and the pairs repeat forever. A general solution formula is simply a way of naming every one of those intersection points with a single expression.

Principal solutions: picking the right ones from one cycle

Before writing the general solution, you need a starting point. A principal solution is a solution that lies in a single, standard interval — typically [0, 2\pi) for sine and cosine, or (-\frac{\pi}{2}, \frac{\pi}{2}) for tangent.

For \sin x = \frac{1}{2}, the principal solutions in [0, 2\pi) are:

The first is the acute angle from the standard table. The second comes from the identity \sin(\pi - \alpha) = \sin \alpha: since \sin \frac{\pi}{6} = \frac{1}{2}, you also have \sin\left(\pi - \frac{\pi}{6}\right) = \sin \frac{5\pi}{6} = \frac{1}{2}.

This is the key geometric insight. On the unit circle, sine is the y-coordinate. Two points on the unit circle share the same y-coordinate: one in the first quadrant, one in the second. Those give the two principal solutions.

For cosine, the picture is similar but reflected. Since \cos x is the x-coordinate, two angles with the same cosine are symmetric about the horizontal axis: one above, one below. Take \cos x = \frac{1}{2}: the principal solutions are x = \frac{\pi}{3} and x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}. The first is in the first quadrant, the second in the fourth. Together, these two points on the unit circle share the same x-coordinate.

More generally, the pair is \alpha and -\alpha (or equivalently \alpha and 2\pi - \alpha). The identity behind this is \cos(-\alpha) = \cos \alpha, which says cosine is an even function — it does not care about the sign of the angle.

For tangent, the picture is different again. Tangent has period \pi (not 2\pi), and within each period there is only one angle with any given tangent value — so there is only one principal solution. For example, \tan x = 1 has the principal solution x = \frac{\pi}{4}, and you do not get a second distinct solution in [0, \pi) because the tangent function is one-to-one on each period.

General solution for \sin x = \sin \alpha

Now for the main result. You want a single formula that captures every solution of \sin x = \sin \alpha, where \alpha is a known angle.

Start with what you know. The equation \sin x = \sin \alpha holds when:

Case 1: x = \alpha + 2n\pi for any integer n. This is just x = \alpha shifted by full periods. The sine function repeats every 2\pi, so adding any integer multiple of 2\pi to a solution gives another solution.

Case 2: x = (\pi - \alpha) + 2n\pi for any integer n. This uses the identity \sin(\pi - \alpha) = \sin \alpha. The angle \pi - \alpha has the same sine as \alpha, and then you shift by full periods.

These two cases look separate, but there is a beautiful way to combine them into one expression. Notice:

• Case 1 gives x = \alpha, \alpha + 2\pi, \alpha + 4\pi, \alpha - 2\pi, \ldots
• Case 2 gives x = \pi - \alpha, 3\pi - \alpha, 5\pi - \alpha, -\pi - \alpha, \ldots

Write n\pi and add (-1)^n \alpha to it. When n is even, (-1)^n = 1, so you get n\pi + \alpha — these are \alpha, 2\pi + \alpha, 4\pi + \alpha, \ldots, which is Case 1. When n is odd, (-1)^n = -1, so you get n\pi - \alpha — these are \pi - \alpha, 3\pi - \alpha, 5\pi - \alpha, \ldots, which is Case 2.

Both cases, merged into one formula:

The (-1)^n is the clever part. It acts as a switch: when n is even, you add \alpha; when n is odd, you subtract \alpha. This single formula produces every solution and no false ones.

Let's verify with a concrete case. Take \sin x = \frac{1}{2}, so \alpha = \frac{\pi}{6}.

• n = 0: x = 0 + (1)\frac{\pi}{6} = \frac{\pi}{6}. Check: \sin \frac{\pi}{6} = \frac{1}{2}. Correct.
• n = 1: x = \pi + (-1)\frac{\pi}{6} = \pi - \frac{\pi}{6} = \frac{5\pi}{6}. Check: \sin \frac{5\pi}{6} = \frac{1}{2}. Correct.
• n = 2: x = 2\pi + (1)\frac{\pi}{6} = \frac{13\pi}{6}. This is \frac{\pi}{6} plus one full cycle. Correct.
• n = -1: x = -\pi + (-1)\frac{\pi}{6} = -\frac{7\pi}{6}. Check: \sin(-\frac{7\pi}{6}) = \sin(\frac{7\pi}{6} \text{ reflected}) = \frac{1}{2}. Correct.

Every integer n gives a valid solution, and every solution of \sin x = \frac{1}{2} appears somewhere in this list.

General solution for \cos x = \cos \alpha

The cosine identity works differently. The two angles with the same cosine are \alpha and -\alpha (since \cos(-\alpha) = \cos \alpha). On the unit circle, these are the two points at the same horizontal distance from the vertical axis — one above, one below.

Adding full periods to each:

• x = \alpha + 2n\pi
• x = -\alpha + 2n\pi

These combine cleanly:

The \pm captures both cases. No (-1)^n trick is needed here — the two families differ only in the sign of \alpha, and the \pm says that directly.

Why is the period 2\pi here but the formula for sine used n\pi? Because the (-1)^n in the sine formula already alternates every \pi, effectively building in the half-period behaviour. The cosine formula does not have that alternation, so it uses the full period 2n\pi explicitly.

General solution for \tan x = \tan \alpha

Tangent has the simplest general solution because tangent has period \pi and is one-to-one within each period.

If \tan x = \tan \alpha, then x and \alpha differ by an integer multiple of \pi:

No \pm, no (-1)^n. Just n\pi + \alpha. The reason is that tangent, unlike sine and cosine, gives each value exactly once per period. There is no "second angle in the period with the same tangent" — the two angles with the same tangent are exactly \pi apart, and the n\pi already accounts for that.

Here is a side-by-side comparison of the three tangent curves, showing how \tan x = 1 produces one solution per period:

The general technique: reduce, then apply

Most trigonometric equations are not handed to you in the form \sin x = \sin \alpha. They come in messier shapes: 2\sin^2 x - 3\sin x + 1 = 0, or \cos 2x = \cos x, or \sin x + \cos x = 1. The technique is always the same:

1. Simplify using identities until the equation involves a single trig function, or takes the form \text{trig}(x) = \text{trig}(\alpha).
2. Reduce to one of the three standard types: \sin x = \sin \alpha, \cos x = \cos \alpha, or \tan x = \tan \alpha.
3. Apply the general solution formula.

Here is the trick that makes step 1 work: many trig equations are quadratics in disguise. If you see \sin^2 x, treat \sin x as a single variable — say, t = \sin x — and solve the resulting algebraic equation for t. Then go back and solve \sin x = t using the general solution.

Special values that come up constantly

Before doing worked examples, here is a reference you will reach for every time:

Equation	General solution
\sin x = 0	x = n\pi
\cos x = 0	x = (2n+1)\frac{\pi}{2}
\tan x = 0	x = n\pi
\sin x = 1	x = 2n\pi + \frac{\pi}{2}
\sin x = -1	x = 2n\pi - \frac{\pi}{2}
\cos x = 1	x = 2n\pi
\cos x = -1	x = (2n+1)\pi

These are just special cases of the general solution formulas you derived above, with \alpha set to the appropriate standard angle. For instance, \sin x = 0 means \sin x = \sin 0, so x = n\pi + (-1)^n \cdot 0 = n\pi.

Common confusions

A few mistakes that come up in almost every exam hall.

• Forgetting the (-1)^n in the sine formula. Students write x = n\pi + \alpha for \sin x = \sin \alpha. That gives only half the solutions — the ones from Case 1. The (-1)^n alternation is what brings in the other half (the \pi - \alpha family). Without it, you are missing every other solution.

• Using degrees inside the general solution. The formulas x = n\pi + (-1)^n \alpha and x = 2n\pi \pm \alpha assume radians. If you work in degrees, replace \pi with 180°: x = n \cdot 180° + (-1)^n \alpha. Mixing degrees and radians in the same formula is a guaranteed error.

• Not checking whether |\text{value}| \leq 1 for sine and cosine. If your quadratic gives \sin x = 3, that has no solution — sine never exceeds 1. Always check the range before writing the general solution.

• Dividing by a trig function. If \sin x \cdot \cos x = \sin x, do not divide by \sin x — you will lose the solutions where \sin x = 0. Instead, bring everything to one side and factor: \sin x (\cos x - 1) = 0.

• Confusing "\sin x = 0" with "x = 0." The equation \sin x = 0 has infinitely many solutions: x = 0, \pm\pi, \pm 2\pi, \ldots The general solution is x = n\pi.

Going deeper

Equations that reduce to a\sin x + b\cos x = c

A frequently tested type is the linear combination a\sin x + b\cos x = c, where a, b, c are constants. The standard technique converts the left side into a single sine (or cosine) using the auxiliary angle method.

Write a\sin x + b\cos x = R\sin(x + \phi), where R = \sqrt{a^2 + b^2} and \tan \phi = \frac{b}{a}.

To see why, expand R\sin(x + \phi) = R\sin x \cos \phi + R\cos x \sin \phi. Matching coefficients with a\sin x + b\cos x gives R\cos \phi = a and R\sin \phi = b. Squaring and adding: R^2 = a^2 + b^2, so R = \sqrt{a^2 + b^2}. Dividing: \tan \phi = b/a.

The equation a\sin x + b\cos x = c becomes R\sin(x + \phi) = c, i.e., \sin(x + \phi) = \frac{c}{R}.

This has solutions only when \left|\frac{c}{R}\right| \leq 1, i.e., when c^2 \leq a^2 + b^2. If that condition holds, you write \frac{c}{R} = \sin \beta for some angle \beta, and apply the general solution for sine:

As a quick example: solve \sin x + \cos x = 1.

Here a = 1, b = 1, c = 1. So R = \sqrt{2}, and \tan \phi = 1, giving \phi = \frac{\pi}{4}.

The equation becomes \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = 1, so \sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}.

General solution: x + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}.

So x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}.

When n is even: x = n\pi + \frac{\pi}{4} - \frac{\pi}{4} = n\pi (i.e., x = 0, 2\pi, 4\pi, \ldots). When n is odd: x = n\pi - \frac{\pi}{4} - \frac{\pi}{4} = n\pi - \frac{\pi}{2} (i.e., x = \frac{\pi}{2}, \frac{5\pi}{2}, \ldots).

You can verify: \sin 0 + \cos 0 = 0 + 1 = 1 and \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1. Both check out.

The number of solutions in a given interval

Competitive exams often ask: "How many solutions does the equation have in [0, 2\pi)?" To answer this, write the general solution, then substitute n = 0, 1, 2, \ldots (and n = -1, -2, \ldots) and check which values of x fall inside the interval. The graph is your friend here — plotting the two sides of the equation and counting intersection points is the fastest visual check.

Here is a quick example. The equation \sin x = \frac{1}{2} has general solution x = n\pi + (-1)^n \frac{\pi}{6}.

• n = 0: x = \frac{\pi}{6} \approx 0.524. Inside [0, 2\pi)? Yes.
• n = 1: x = \frac{5\pi}{6} \approx 2.618. Inside? Yes.
• n = 2: x = \frac{13\pi}{6} \approx 6.807. Inside? No — 2\pi \approx 6.283 and \frac{13\pi}{6} > 2\pi.
• n = -1: x = -\frac{7\pi}{6} < 0. No.

So there are exactly two solutions in [0, 2\pi). This matches the graph: the horizontal line y = \frac{1}{2} crosses the sine curve twice in each complete period.

A note on Bhaskara II's approach

Indian mathematicians worked with trigonometric equations centuries before the modern notation existed. Bhaskara II, in the Siddhanta Shiromani (12th century), systematically solved equations involving jya (sine) and kotijya (cosine) to compute astronomical quantities — the positions of planets require solving exactly the kinds of equations you have just learned. The general solution formulas were implicit in his methods, even though the notation n\pi + (-1)^n\alpha came much later.

Where this leads next

You now have the tools to solve any trigonometric equation that reduces to a single trig function. The natural next steps:

• Trigonometric Equations — Advanced — equations involving multiple angles, systems of trig equations, and parametric solutions.
• Inverse Trigonometric Functions — the formal framework for picking the "right" single solution, which is what principal value branches do.
• Trigonometric Identities — the toolkit of identities you use to simplify equations before solving.
• Multiple Angles — double-angle, triple-angle, and half-angle formulas that appear frequently in trig equations.
• Compound Angles — the addition formulas that underpin the auxiliary angle method.