Trigonometric Equations Worked ExamplesIGCSE Maths 0580

Make $\sin x$ the subject.

Use the exact value you know: $\sin 30^{\circ} = \tfrac{1}{2}$.

Sine is positive in the first and second quadrants, and the sine curve is symmetrical about $x = 90^{\circ}$. The second solution is the supplement.

Both lie in the interval, and there are no further solutions before $360^{\circ}$, where $\sin x$ is negative.

The inverse cosine of a negative value returns an obtuse angle directly.

The cosine curve is symmetrical about $x = 180^{\circ}$, and it is below the axis between $90^{\circ}$ and $270^{\circ}$. The second solution is found by subtracting from $360^{\circ}$.

The calculator gives the first-quadrant solution.

Tangent repeats every $180^{\circ}$, not $360^{\circ}$, so add $180^{\circ}$ for the next solution.

Adding another $180^{\circ}$ exceeds $360^{\circ}$, so stop.

Isolate the trigonometric function before doing anything else.

Take the inverse cosine.

Second solution for cosine.

This is a quadratic in $\sin x$. Substituting $s = \sin x$ makes that visible.

Factorise as an ordinary quadratic.

Solve $\sin x = 1$. This occurs once in the interval, at the maximum of the sine curve.

Solve $\sin x = -\tfrac{1}{2}$. Sine is negative in the third and fourth quadrants; the reference angle is $30^{\circ}$.

(a) Subtract $4$ from both sides, then divide by $-7$. Dividing by a negative number changes the sign of the right-hand side.

(b) The cosine curve is below the $x$-axis only between its crossings at $90^{\circ}$ and $270^{\circ}$. Since $-\tfrac{5}{7}$ is between $-1$ and $0$, the line $y = -\tfrac{5}{7}$ cuts the curve twice, once either side of the minimum at $180^{\circ}$.

(c) The inverse cosine of a negative value gives the first solution directly.

The curve is symmetrical about $x = 180^{\circ}$, so the second solution is $360^{\circ}$ minus the first.

Both values are between $90^{\circ}$ and $270^{\circ}$, as the sketch predicted. Check: $4 - 7\cos 135.58^{\circ} = 4 + 5.00 = 9.00$. ✓

(a) Rearrange to make $\tan x$ the subject.

So the solutions are where $y = \tan x$ meets the line $y = -\tfrac{4}{3}$. The first branch ($0^{\circ}$ to $90^{\circ}$) is never negative, so the line misses it. It crosses the branch after each asymptote once: two solutions, one between $90^{\circ}$ and $180^{\circ}$ and one between $270^{\circ}$ and $360^{\circ}$.

(b) The calculator returns a negative angle, which is outside the interval. Do not stop here — use it to generate the others.

The tangent graph repeats every $180^{\circ}$. Add $180^{\circ}$ once, then again.

Check against the sketch: $126.9^{\circ}$ is between $90^{\circ}$ and $180^{\circ}$, and $306.9^{\circ}$ is between $270^{\circ}$ and $360^{\circ}$. ✓

(a) Taking the square root of $\cos^2 x = \tfrac{1}{4}$ gives two values, because $\left(-\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ as well. Leon only used the positive root, so he lost every solution of $\cos x = -\tfrac{1}{2}$.

(b) The sketch shows each dashed line cutting the curve twice, so expect four solutions. For $\cos x = \tfrac{1}{2}$, Leon's two answers are correct: $\cos 60^{\circ} = \tfrac{1}{2}$, and the curve is symmetrical about $x = 180^{\circ}$.

For $\cos x = -\tfrac{1}{2}$ the crossings are below the axis, either side of the minimum at $180^{\circ}$. They are $60^{\circ}$ either side of it, using the same reference angle.

(c) The same steps give $\sin x = \pm\tfrac{1}{2}$. On the graph of $y = \sin x$ the lines $y = \tfrac{1}{2}$ and $y = -\tfrac{1}{2}$ each cross the curve twice (neither touches a maximum or minimum), so there are again four solutions: $30^{\circ}$, $150^{\circ}$, $210^{\circ}$ and $330^{\circ}$.

Make $\sin x$ the subject, so the equation can be read from the sketch. The solutions are where the horizontal line $y = \frac{k-1}{2}$ meets the curve.

(a) Only one horizontal line meets the curve three times: the $x$-axis, $y = 0$, at $0^{\circ}$, $180^{\circ}$ and $360^{\circ}$ (the interval includes both end points).

(b) A line meets the curve exactly once only where it just touches the maximum or the minimum.

(c) Any other line strictly between $y = -1$ and $y = 1$ cuts the curve twice — but $y = 0$ must be left out, because it gives three. Then convert back to $k$ by multiplying by $2$ and adding $1$.

(d) Here $k = 0.2$, which is in the range $-1 < k < 1$, so expect two solutions, both where the curve is below the axis (between $180^{\circ}$ and $360^{\circ}$).

The calculator gives $\sin^{-1}(-0.4) = -23.578\ldots^{\circ}$, outside the interval. Use the reference angle $23.578\ldots^{\circ}$: sine is negative in the third and fourth quadrants, so add it to $180^{\circ}$ and subtract it from $360^{\circ}$.

(a) Subtract $7$ from both sides first. The $7$ is not a multiple of $\tan x$, so it cannot be combined with the $4\tan x$.

The calculator gives a negative angle, outside the interval.

The tangent graph repeats every $180^{\circ}$, so add $180^{\circ}$ once and then again to land inside the interval.

(b) An angle between two crossing lines is less than $180^{\circ}$, and $y$ is obtuse, so $y = 143.130\ldots^{\circ}$. The angle next to it on the same straight line makes $180^{\circ}$ with it.

Vertically opposite angles are equal, so the four angles are $y$, its neighbour, and a copy of each. Check: $2(143.1^{\circ}) + 2(36.9^{\circ}) = 360^{\circ}$. ✓

Subtract $3$ from both sides, then divide by $10$.

A sketch of $y = \sin x$ shows the line $y = -0.7$ crosses the curve twice, both between $180^{\circ}$ and $360^{\circ}$. The calculator value is outside the interval, but its size is the reference angle.

Use the symmetry of the sine curve about $x = 270^{\circ}$: one solution is $44.427\ldots^{\circ}$ after $180^{\circ}$, the other $44.427\ldots^{\circ}$ before $360^{\circ}$.