PT5 - Solving trigonometric equations

Find every solution of a trigonometric equation in a stated interval, rather than stopping at the first calculator answer. You will use the symmetry and periods of sine, cosine and tangent, adjust intervals when the angle changes, and solve simple equations using the two trigonometric identities you already know.

Finding every sine solution

Why does $\sin2x=0.6$ have four solutions between $0^\circ$ and $360^\circ$ , when your calculator gives only one inverse-sine answer? The key is that the calculator chooses one angle, while the rotating point can reach the same height repeatedly. We will find all four before the lesson ends.

You already know

Sine is the vertical coordinate on the unit circle, cosine is the horizontal coordinate and tangent is their quotient. Sine and cosine repeat every $360^\circ$ ; tangent repeats every $180^\circ$ . Work in degrees.

For $\sin x=0.5$ , the point has height $0.5$ twice in one turn: at $30^\circ$ and $150^\circ$ . Both are solutions. Inverse sine returns $30^\circ$ , called its principal value, because its output is restricted to $-90^\circ\le x\le90^\circ$ . It has not searched the interval in your question.

For a negative sine, use the acute reference angle for the size, then put the point below the horizontal axis. For example, $\sin x=-0.5$ gives $210^\circ$ and $330^\circ$ in one turn. These are reflections of the positive-height points.

Angles with the same sine

If $\sin x=\sin\alpha$ , then

x=\alpha+360^\circ n \qquad\text{or}\qquad x=180^\circ-\alpha+360^\circ n

Here $n$ is any integer. Keep only angles in the requested interval and list repeated angles once. No solution exists for $\sin x=k$ if $|k|>1$ .
Not on the formulae sheet: learn it.

The graph or circle explains the two families; the interval tells you which members to keep. You can use a positive reference angle instead of the principal value if you assign the quadrants carefully.

Worked example 1: a positive sine

Question. Solve $\sin x=0.4$ for $0^\circ\le x\le360^\circ$ , giving answers to 1 decimal place.

Plan. Find the acute reference angle, reflect it into quadrant II, and check the endpoints.

\begin{aligned} \alpha&=\sin^{-1}(0.4)=23.5781\ldots^\circ &&\text{principal angle}\\ x_1&=\alpha &&\text{quadrant I}\\ x_2&=180^\circ-\alpha=156.4218\ldots^\circ &&\text{quadrant II} \end{aligned}

Why these quadrants? A positive sine means a point above the horizontal axis. There are exactly two such points at this height in one turn.

Answer. $x=23.6^\circ$ or $156.4^\circ$ .

Check. Substitution of the unrounded angles gives sine $0.4$ in each case. The endpoints have sine $0$ , so neither is a solution.

The next example changes only the sign of the right-hand side. The reference angle stays the same; the solutions move below the axis.

Worked example 2: a negative sine

Question. Solve $\sin x=-0.4$ for $0^\circ\le x\le360^\circ$ , giving answers to 1 decimal place.

Plan. Use the same acute reference angle, then locate the two negative-height points.

\begin{aligned} \alpha&=\sin^{-1}(0.4)=23.5781\ldots^\circ &&\text{reference angle}\\ x_1&=180^\circ+\alpha=203.5781\ldots^\circ &&\text{quadrant III}\\ x_2&=360^\circ-\alpha=336.4218\ldots^\circ &&\text{quadrant IV} \end{aligned}

Why not keep the calculator's $-23.5781\ldots^\circ$ ? Its sine is correct, but it lies outside the requested interval. Adding a full turn gives the quadrant IV solution.

Answer. $x=203.6^\circ$ or $336.4^\circ$ .

Check. Both angles lie below the horizontal axis and have reference angle $\alpha$ , so both have sine $-0.4$ . The endpoints again fail.

At the top or bottom of the circle, the two families coincide: $\sin x=1$ has only $90^\circ$ in one turn. At height zero, check both ends of a closed interval: $\sin x=0$ on $[0^\circ, 360^\circ]$ has the three distinct angle values $0^\circ$ , $180^\circ$ and $360^\circ$ . The endpoints represent the same point but different allowed angles.

Watch out

Wrong: $\sin x=-0.4\Rightarrow x=-23.6^\circ$ only. An inverse function gives one principal value. Right: generate the other angles with the same sine and apply the stated interval.

Try a new negative height. Give both solutions, not just the principal value.

Finding every cosine solution

Cosine equations ask for a horizontal coordinate, so reflect the point in the horizontal axis to find the second solution.

The principal value of inverse cosine lies between $0^\circ$ and $180^\circ$ . For example, $\cos x=-0.5$ gives principal angle $120^\circ$ , and reflection gives $360^\circ-120^\circ=240^\circ$ . Subtracting from $180^\circ$ would reverse the horizontal coordinate instead of preserving it.

Angles with the same cosine

If $\cos x=\cos\alpha$ , then

x=\alpha+360^\circ n \qquad\text{or}\qquad x=-\alpha+360^\circ n

Here $n$ is any integer. Retain only angles in the requested interval and list duplicates once. No solution exists for $\cos x=k$ if $|k|>1$ .
Not on the formulae sheet: learn it.

Within one turn, the second angle is usually most conveniently written $360^\circ-\alpha$ .

Worked example 3: a negative cosine

Question. Solve $\cos x=-0.3$ for $0^\circ\le x\le360^\circ$ , giving answers to 1 decimal place.

Plan. Find the principal cosine angle and reflect it across the horizontal axis.

\begin{aligned} \alpha&=\cos^{-1}(-0.3)=107.4576\ldots^\circ &&\text{principal angle in quadrant II}\\ x_1&=\alpha &&\text{first solution}\\ x_2&=360^\circ-\alpha=252.5423\ldots^\circ &&\text{reflection into quadrant III} \end{aligned}

Why reflect this way? Both points must be left of the vertical axis, because the cosine is negative.

Answer. $x=107.5^\circ$ or $252.5^\circ$ .

Check. Both unrounded angles have cosine $-0.3$ . The endpoints have cosine $1$ , so neither is included.

The next example changes only the sign of the required horizontal coordinate.

Worked example 4: a positive cosine

Question. Solve $\cos x=0.3$ for $0^\circ\le x\le360^\circ$ , giving answers to 1 decimal place.

Plan. Find the principal angle, then use its reflection in quadrant IV.

\begin{aligned} \alpha&=\cos^{-1}(0.3)=72.5423\ldots^\circ &&\text{principal angle in quadrant I}\\ x_1&=\alpha &&\text{first solution}\\ x_2&=360^\circ-\alpha=287.4576\ldots^\circ &&\text{reflection into quadrant IV} \end{aligned}

Why not use $180^\circ-\alpha$ ? That point is in quadrant II and has cosine $-0.3$ , the wrong sign.

Answer. $x=72.5^\circ$ or $287.5^\circ$ .

Check. Both points are right of the vertical axis. Cosine is $1$ at each endpoint, so there are no endpoint solutions.

Cosine's extreme values need care too. On this closed interval, $\cos x=1$ has $0^\circ$ and $360^\circ$ , whereas $\cos x=-1$ has just $180^\circ$ .

Watch out

Wrong: $\cos x=0.3\Rightarrow x=72.5^\circ,\ 107.5^\circ$ . Those angles have opposite cosines. Right: use $360^\circ-72.5423\ldots^\circ$ , giving $287.5^\circ$ .

Use the horizontal-coordinate reflection with a different value.

Using tangent's half-turn period

For tangent, move through half-turns rather than using the sine or cosine reflection rules.

Opposite points on the unit circle have both coordinates reversed, so their quotient is unchanged. Inverse tangent returns its principal angle strictly between $-90^\circ$ and $90^\circ$ . Every further solution is obtained by adding or subtracting $180^\circ$ .

Angles with the same tangent

If $\tan x=k$ and $\alpha=\tan^{-1}k$ , then

x=\alpha+180^\circ n

Here $n$ is any integer. Tangent is undefined at $90^\circ+180^\circ n$ ; those angles cannot be solutions. Every real $k$ is a possible tangent value.
Not on the formulae sheet: learn it.

Choose enough integers to cover the interval, then discard the values outside it. A negative principal angle is useful; there is no need to force it to be positive before starting.

Worked example 5: a positive tangent

Question. Solve $\tan x=1.5$ for $0^\circ\le x\le360^\circ$ , giving answers to 1 decimal place.

Plan. Find the principal angle and add half-turns until the interval is covered.

\begin{aligned} \alpha&=\tan^{-1}(1.5)=56.3099\ldots^\circ &&\text{principal angle}\\ \alpha+180^\circ&=236.3099\ldots^\circ &&\text{next solution}\\ \alpha+360^\circ&=416.3099\ldots^\circ &&\text{outside the interval} \end{aligned}

Why add $180^\circ$ ? Tangent repeats after a half-turn. Adding only full turns would miss a solution.

Answer. $x=56.3^\circ$ or $236.3^\circ$ .

Check. Tangent is positive in quadrants I and III, matching both answers. Its value is $0$ at the endpoints.

The next example changes only the sign of the tangent, so the principal value becomes negative.

Worked example 6: starting with a negative angle

Question. Solve $\tan x=-1.5$ for $0^\circ\le x\le360^\circ$ , giving answers to 1 decimal place.

Plan. Retain the negative principal value and add successive half-turns.

\begin{aligned} \alpha&=\tan^{-1}(-1.5)=-56.3099\ldots^\circ &&\text{outside the interval}\\ \alpha+180^\circ&=123.6900\ldots^\circ &&\text{quadrant II solution}\\ \alpha+360^\circ&=303.6900\ldots^\circ &&\text{quadrant IV solution}\\ \alpha+540^\circ&=483.6900\ldots^\circ &&\text{outside the interval} \end{aligned}

Why keep adding from the negative value? It preserves the exact half-turn spacing and avoids inventing a reflection rule.

Answer. $x=123.7^\circ$ or $303.7^\circ$ .

Check. Both angles have tangent $-1.5$ before rounding and lie in the negative-tangent quadrants. Neither endpoint is a root.

Each tangent branch contributes at most one root, because it increases continuously between its asymptotes.

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Numerical methods

PT5 - Solving trigonometric equations

Finding every sine solution

Finding every cosine solution

Using tangent's half-turn period

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