Solving Trig Equations

Ask a calculator to solve \sin\theta = 0.5 and it answers 30^\circ, full stop. Tidy — and wrong, or at least badly incomplete. Because \theta = 150^\circ works too (\sin 150^\circ = 0.5). So does 390^\circ, and 510^\circ, and 750^\circ, marching on forever in both directions.

A trig equation doesn't usually have one answer — it has infinitely many, because the sine and cosine curves repeat. That's not a nuisance; it's the whole point. These functions describe things that recur — tides, seasons, the swing of a pendulum, the voltage in a mains socket — so "when does \sin\theta = 0.5?" is really asking "when does this repeating thing hit that value?", and the honest answer is "again and again". Solving a trig equation means finding all the solutions inside a stated range.

The method: principal value, then symmetry

There are just two moves.

Get the principal value. Use the inverse function — \theta = \sin^{-1}(k), or \cos^{-1}, \tan^{-1} — to get the one angle the calculator will give you.
Use the curve's symmetry to find the rest of the solutions in the range. Sine is positive in the second quadrant as well as the first, so if \theta works, so does 180^\circ - \theta. Then add 360^\circ to any solution to step to the next one, until you leave the range.

The symmetry partner depends on which ratio you're solving:

For sine, the second solution is 180^\circ - \theta.
For cosine, the second solution is 360^\circ - \theta.
Tangent repeats every 180^\circ, so the next solution is \theta + 180^\circ.

Find the principal value \theta = \sin^{-1}(k) (or \cos^{-1}, \tan^{-1}) with the inverse function.
Use the graph's symmetry to read off the other solutions inside the range, then add 360^\circ to reach further ones.
\sin\theta = k \;\Rightarrow\; \{\,\theta,\; 180^\circ - \theta,\; \theta+360^\circ,\;\dots\,\}
\cos\theta = k \;\Rightarrow\; \{\,\theta,\; 360^\circ - \theta,\; \theta+360^\circ,\;\dots\,\}
\tan\theta = k \;\Rightarrow\; \{\,\theta,\; \theta + 180^\circ,\;\dots\,\}

Worked example 1 — two crossings

Solve \sin\theta = 0.5 for 0^\circ \le \theta \le 360^\circ.

Principal value: \theta = \sin^{-1}(0.5) = 30^\circ. Symmetry: for sine the partner is 180^\circ - 30^\circ = 150^\circ. Adding another 360^\circ would give 390^\circ, which is outside the range — so we stop. The solutions are \theta = 30^\circ and \theta = 150^\circ.

The picture below shows why: y = \sin(x^\circ) meets the flat line y = 0.5 at exactly those two points between 0^\circ and 360^\circ. Solving the equation is finding where the wave crosses the line.

Worked example 2 — a cosine equation

Solve \cos\theta = 0.5 for 0^\circ \le \theta \le 360^\circ.

Principal value: \theta = \cos^{-1}(0.5) = 60^\circ. Symmetry: for cosine the partner is 360^\circ - 60^\circ = 300^\circ. Both lie in the range, so the solutions are \theta = 60^\circ and \theta = 300^\circ. Notice the rule changed with the function — sine used 180^\circ - \theta, cosine uses 360^\circ - \theta. Reach for the wrong partner and you'll get the wrong second answer.

Worked example 3 — rearrange first

Solve 2\sin\theta - 1 = 0 for 0^\circ \le \theta \le 360^\circ.

Many exam equations hide the trig ratio inside some algebra. Get the ratio on its own first, then run the two-step method:

2\sin\theta - 1 = 0 \;\Rightarrow\; 2\sin\theta = 1 \;\Rightarrow\; \sin\theta = \tfrac{1}{2}.

Now it's the equation we already solved: \theta = 30^\circ and \theta = 150^\circ. The rearranging is just clearing the path so the principal-value-then-symmetry routine can do its work.

Worked example 4 — a wider range

Solve \sin\theta = 0.5 again, but now for 0^\circ \le \theta \le 720^\circ.

Nothing about the method changes — only the range is bigger, so there is more room for solutions. Start with the two we already know, 30^\circ and 150^\circ, then keep adding 360^\circ until you run off the end of the range:

30^\circ,\ \ 150^\circ,\ \ 30^\circ + 360^\circ = 390^\circ,\ \ 150^\circ + 360^\circ = 510^\circ.

Adding another 360^\circ would overshoot 720^\circ, so we stop. Four solutions this time, not two — the equation didn't change, the interval did. This is exactly why re-reading the range is a step, not an afterthought.

The mistakes that lose the most marks on this topic:

The calculator gives you only ONE solution. \sin^{-1}, \cos^{-1} and \tan^{-1} each hand back a single principal value. If you write it down and stop, you'll drop every other solution in the range. The symmetry step is not optional — it's where most of the answers live.
The interval decides how many answers there are. The same equation has two solutions in 0^\circ–360^\circ, four in 0^\circ–720^\circ, and one (or zero) in a narrow range. Always re-read the range, and sanity-check how many crossings it should contain before you commit.
Degrees or radians? If the range is written in radians (like 0 \le \theta \le 2\pi) your calculator must be in radian mode, and your symmetry rules become \pi - \theta and 2\pi - \theta. Mixing the two modes is a silent, marks-shredding error.

Everything that goes round and round or up and down is modelled by a sine or cosine wave — and every time you ask "when does it reach a certain value?" you are solving a trig equation. When is a Ferris-wheel car exactly 20\,\text{m} above the ground? When does the tide reach the harbour wall? When does the mains voltage peak, sixty times a second? When, across the year, are there exactly twelve hours of daylight?

Every one of those questions has more than one answer — the car comes back to that height on every rotation, the tide returns twice a day, the voltage peaks over and over. The "infinitely many solutions" that felt like a technicality are simply the mathematics being honest: repeating events recur. You'll see exactly why the graph has this endless symmetry when you meet the properties of the sine curve.