Trig Equations: When sin x Is Negative, Where Are the Two Angles?

The question

Solve the equation

5\sin x + 3 = 2\sin x + 1

for $0^\circ \le x \le 360^\circ$ . Give your answers correct to 1 decimal place. [3 marks]

Cambridge IGCSE note: solving trigonometric equations for $0^\circ \le x \le 360^\circ$ is Extended content in 0580.

The common wrong answer

The first step is usually fine. Collect the $\sin x$ terms on one side:

3\sin x = -2 \quad\Rightarrow\quad \sin x = -\frac{2}{3}

Then the calculator gives

x = \sin^{-1}\left(-\frac{2}{3}\right) = -41.8^\circ

That isn’t in the range $0^\circ$ to $360^\circ$ , so many students quietly drop the minus sign and use $41.8^\circ$ . Then they remember that sine has a second solution at $180^\circ$ minus the angle, and write

x = 41.8^\circ \quad\text{or}\quad x = 180^\circ - 41.8^\circ = 138.2^\circ

Both of these are wrong. Check them on the calculator: $\sin 41.8^\circ = 0.667$ and $\sin 138.2^\circ = 0.667$ . They solve $\sin x = +\frac{2}{3}$ , not $\sin x = -\frac{2}{3}$ . Angles between $0^\circ$ and $180^\circ$ always have a positive sine, so neither answer can be right.

Cambridge examiners reported this same mistake, with different numbers, as a very common incorrect answer in a recent paper. Two other slips they saw: using $180^\circ$ minus the angle instead of $360^\circ$ minus it, and adding the angle to $360^\circ$ , which gives an answer above $360^\circ$ .

The correct working

Rearrange to get $\sin x$ on its own:

3\sin x = -2 \quad\Rightarrow\quad \sin x = -\frac{2}{3}

Find the reference angle (the acute angle) using the positive value:

\sin^{-1}\left(\frac{2}{3}\right) = 41.81^\circ

Sine is negative in the third and fourth quadrants, so:

\begin{aligned} \text{Quadrant III:}\quad x &= 180^\circ + 41.81^\circ = 221.81^\circ \\ \text{Quadrant IV:}\quad x &= 360^\circ - 41.81^\circ = 318.19^\circ \end{aligned}

So $x = 221.8^\circ$ or $x = 318.2^\circ$ .

The graph shows why. The line $y = -\frac{2}{3}$ crosses the sine curve exactly twice between $0^\circ$ and $360^\circ$ , and both crossings are in the half of the curve below the $x$ -axis.

Graph of y = sin x for x from 0 to 360 degrees. The line y = −2/3 crosses the curve at x = 221.8 degrees and x = 318.2 degrees. A dashed line y = 2/3 crosses the curve at 41.8 and 138.2 degrees, the wrong answers. — y = sin x meets y = −2/3 at 221.8° and 318.2°. The wrong answers, 41.8° and 138.2°, lie on y = 2/3.

Check: $\sin 221.8^\circ = -0.667$ and $\sin 318.2^\circ = -0.667$ , and both angles are between $0^\circ$ and $360^\circ$ . ✓

Why it works

The calculator only ever gives one angle, and for $\sin^{-1}$ it is always between $-90^\circ$ and $90^\circ$ . Every other solution has to come from the shape of the graph (or the quadrants), not from the calculator.

Working with the reference angle keeps the two jobs separate. The size of the number, $\frac{2}{3}$ , tells you the reference angle, $41.81^\circ$ . The sign tells you which quadrants to use. Sine is positive in quadrants I and II and negative in quadrants III and IV, so a negative value sends you to $180^\circ + 41.81^\circ$ and $360^\circ - 41.81^\circ$ .

The $-41.8^\circ$ from the calculator isn’t useless, either. Going $41.8^\circ$ clockwise from $0^\circ$ lands at the same place as $360^\circ - 41.8^\circ = 318.2^\circ$ . That’s one of the answers; the other is in quadrant III.

How to spot it next time

If $\sin x$ is negative, both answers must be between $180^\circ$ and $360^\circ$ . An answer below $180^\circ$ is a warning sign.
Find the reference angle from the positive value, then use the sign to pick the quadrants: III is $180^\circ + \text{angle}$ , IV is $360^\circ - \text{angle}$ .
Put each answer back into the calculator and check you get the value you started with, sign included.