Solving Trigonometric Equations

Trigonometry connects the angles of a triangle to the lengths of its sides. Three ratios named sine, cosine, and tangent do most of the work.

Solving Trigonometric Equations: the central idea

A trigonometric equation usually represents infinitely many angles. Solving requires a reference angle, quadrant signs, and either an interval list or a general periodic form.

Words you need

Reference angle

A reference angle is the sharp positive angle between the terminal side and the $x$-axis, and in a trig equation it is the first thing you find, because it fixes the size of every solution before signs are considered.

Principal value

The principal value is the single angle an inverse function returns, taken from its restricted range: $[-\pi/2,\pi/2]$ for arcsine, $[0,\pi]$ for arccosine, and $(-\pi/2,\pi/2)$ for arctangent.

Period

The period is how far you travel along the input before the function repeats, which is $2\pi$ for sine, cosine, cosecant, and secant, and $\pi$ for tangent and cotangent.

General solution

A general solution is the formula that captures every answer at once by adding a whole-number multiple of the period, such as $x=\pi/6+2\pi n$.

Extraneous solution

An extraneous solution is a value that survives your algebra but fails the original equation, and it usually appears after squaring both sides, which is why every candidate must be substituted back.

Interval notation

Interval notation such as $[0,2\pi)$ names the window of answers you must report, and the square bracket includes $0$ while the round bracket excludes $2\pi$, so a solution at $0$ counts but one at $2\pi$ does not.

What to know before this lesson

Know exact unit-circle values, inverse trig functions, algebraic equation solving, and periodicity.

If one of those prerequisites is uncertain, use the Trigonometry subject guide to locate the earlier concept before memorizing a procedure.

Trigonometric equations: a worked example

Every step, with the arithmetic

1. Step 1 - Spot the hidden quadraticSolve $2\cos^2x+\cos x-1=0$ on $[0,2\pi)$. Let $u=\cos x$, giving $2u^2+u-1=0$.
2. Step 2 - Factor it$2u^2+u-1=(2u-1)(u+1)=0$
3. Step 3 - Solve for $u$$2u-1=0$ gives $u=\tfrac12$, and $u+1=0$ gives $u=-1$.
4. Step 4 - Translate back to cosine$\cos x=\tfrac12$ or $\cos x=-1$
5. Step 5 - Handle $\cos x=\tfrac12$The reference angle is $\pi/3$. Cosine is positive in quadrants I and IV, so $x=\dfrac{\pi}{3}$ and $x=2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}$.
6. Step 6 - Handle $\cos x=-1$Cosine equals $-1$ only at the far-left point $(-1,0)$ of the unit circle, so $x=\pi$. There is no second angle here.
7. Step 7 - Collect the interval answers$x=\dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{3}$
8. Step 8 - Write the general solutions$x=\pm\dfrac{\pi}{3}+2\pi n$ and $x=\pi+2\pi n$, for every whole number $n$.

For $\sin x=1/2$ on $[0,2\pi)$, the reference angle is $\pi/6$ and sine is positive in quadrants I and II, giving $\pi/6$ and $5\pi/6$.

Where each equation's solutions sit, and how far apart the repeats are

The step-by-step method for trigonometric equations

1. Isolate one trigonometric function and find a reference angle from an exact value or inverse function.
2. Choose every quadrant where the function has the required sign.
3. List solutions in the requested interval or attach the correct period to a general solution, then substitute back.

How to check your answer

Evaluate sine at both listed angles and scan one full period to confirm there are no omitted intersections.

Confirm the quadrant and sign, convert degrees and radians consistently, and substitute solutions into the original interval because periodic equations usually have more than one answer.

A mistake that changes the mathematics

An inverse-sine result is only the principal value. It does not automatically list the second quadrant or later periods.

Where you will actually use this

Tide tables

A harbour's depth is modelled by $h=5+3\sin\left(\dfrac{\pi t}{6}\right)$ metres, with $t$ in hours. A boat needing $6.5$ metres of water asks when $3\sin(\pi t/6)=1.5$, so $\sin(\pi t/6)=0.5$ and $\pi t/6=\pi/6$ or $5\pi/6$, giving $t=1$ and $t=5$ hours. The boat has a four-hour window.

Daylight hours through the year

Day length in a temperate city follows a sine curve with a period of $365$ days. Asking when daylight first reaches $14$ hours is a trig equation, and the two solutions it returns are the matching dates in spring and late summer.

Electrical engineering

Household voltage is $v=170\sin(120\pi t)$. Finding every instant the voltage crosses a $120$-volt threshold means solving $\sin(120\pi t)=0.706$, and because the period is only $1/60$ of a second, the general solution with $+2\pi n$ is the only practical way to list the crossings.

How trigonometric equations connects to the rest of trigonometry

• The unit circle — Choosing which quadrants to take is pure unit-circle reading, so a memorized circle turns the hardest step of solving into a glance.
• Trigonometric identities — Equations mixing two different functions usually will not budge until an identity rewrites them in one function, such as replacing $\sin^2x$ with $1-\cos^2x$.
• Quadratic formula — Many trig equations are quadratics in disguise; substitute $u=\sin x$ or $u=\cos x$ and the familiar factoring or formula finishes the algebra before you return to angles.

Try a transfer problem

Solve $2\cos^2x-1=0$ on $[0,2\pi)$ and state the corresponding general solutions.

Show the worked answer

Start with $2\cos^2x-1=0$, so $2\cos^2x=1$ and $\cos^2x=\tfrac12$. Taking square roots gives both signs: $\cos x=\pm\dfrac{\sqrt2}{2}$. The reference angle for $\dfrac{\sqrt2}{2}$ is $\dfrac{\pi}{4}$. Cosine is positive in quadrants I and IV, giving $x=\dfrac{\pi}{4}$ and $x=\dfrac{7\pi}{4}$; cosine is negative in quadrants II and III, giving $x=\dfrac{3\pi}{4}$ and $x=\dfrac{5\pi}{4}$. So on $[0,2\pi)$ the four solutions are $\dfrac{\pi}{4},\ \dfrac{3\pi}{4},\ \dfrac{5\pi}{4},\ \dfrac{7\pi}{4}$. These four sit evenly spaced a quarter-turn apart, so all of them collapse into the single general solution $x=\dfrac{\pi}{4}+\dfrac{\pi n}{2}$. Checking $n=0,1,2,3$ regenerates exactly the four angles above.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to trigonometry practice questions.

Questions about trigonometric equations

Why does my calculator give only one answer?

Because inverse trig functions must be functions, so each is restricted to one output per input. Arcsine only returns angles in $[-\pi/2,\pi/2]$, arccosine only in $[0,\pi]$. The calculator is not wrong, it is just handing you the principal value and leaving the rest of the circle to you.

How do I solve something like sin(2x) = 1/2?

Solve for the whole inside piece first, over a stretched interval. If $x$ ranges over $[0,2\pi)$, then $2x$ ranges over $[0,4\pi)$, where $\sin$ equals $\tfrac12$ at $\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{13\pi}{6},\dfrac{17\pi}{6}$. Divide each by $2$ to get $x=\dfrac{\pi}{12},\dfrac{5\pi}{12},\dfrac{13\pi}{12},\dfrac{17\pi}{12}$. Doubling the angle doubles the number of solutions.

Can I divide both sides by sin x?

No, and this is the most common way to lose answers. In $2\sin x\cos x=\sin x$, dividing by $\sin x$ throws away every angle where $\sin x=0$. Move everything to one side and factor instead: $\sin x(2\cos x-1)=0$ gives $x=0,\pi$ from the first factor and $x=\dfrac{\pi}{3},\dfrac{5\pi}{3}$ from the second.

Why is the period of tangent pi but sine 2pi?

Because $\tan x=y/x$ on the unit circle, and rotating a half turn flips the signs of both coordinates. Two negatives divide to a positive, so the ratio comes back to the same value after only $\pi$. Sine is the raw $y$-value, which does change sign after a half turn, so it needs a full $2\pi$ to return.