Quadratic Equations in Sine, Cosine or Tangent

An equation that is quadratic in sin x

Look at the equation $2 \sin^{2}x - \sin x - 1 = 0$, to be solved for $0^\circ \le x < 360^\circ$. The notation $\sin^{2}x$ means $\left(\sin x\right)^{2}$, the sine of x, squared. It does not mean the sine of $x^{2}$.

The equation holds sin x squared, sin x on its own, and a number, and nothing else. That is the shape of a quadratic equation. To see it plainly, write s for sin x. The equation becomes $2s^{2} - s - 1 = 0$, a quadratic in s of the kind you already know how to solve.

The plan has two stages. First solve the quadratic for s. Then put sin x back in place of s, and solve each of the simple equations that come out.

Factor and read off the roots

Factor $2s^{2} - s - 1$. Two numbers that multiply to 2 × (−1) = −2 and add to −1 are −2 and 1, so split the middle term: $2s^{2} - 2s + s - 1 = 2s\left(s - 1\right) + 1\left(s - 1\right) = \left(2s + 1\right)\left(s - 1\right)$.

Check by expanding: $\left(2s + 1\right)\left(s - 1\right) = 2s^{2} - 2s + s - 1 = 2s^{2} - s - 1$, which is the equation.

A product is 0 only when one of its factors is 0. So either 2s + 1 = 0, which gives s = −½, or s − 1 = 0, which gives s = 1. The roots are the values sin x can take: sin x = −½ or sin x = 1.

The quadratic $y = 2c^{2} + 3c - 2$ from the cosine equation below, with c standing for cos x. It crosses the axis at c = ½ and c = −2. A cosine lies between the dashed lines c = −1 and c = 1, so only the root ½ can be a value of cos x.

Put sin x back: the root −½

Take the first root, sin x = −½. The inverse sine gives $\sin^{-1}\left(-\frac{1}{2}\right) = -30^\circ$, which is outside the range, so add a full turn: −30° + 360° = 330°. The sine pairing gives the second angle, 180° − (−30°) = 210°.

So sin x = −½ has two solutions in the range: x = 210° and x = 330°. On the graph of y = sin x, the horizontal line y = −½ crosses the wave twice in one turn, once on the way down to the trough at 270° and once on the way back up.

y = sin x over one turn, with one square across for every 90°. The line y = −½ meets the wave at 210° and 330°. The line y = 1 only touches it, at the peak at 90°.

The root 1, and the whole answer

Now the second root, sin x = 1. The sine reaches 1 only at the top of the circle, so in one turn there is a single solution, x = 90°. The pairing 180° − 90° gives 90° again: the two solutions have merged into one.

The answer to the original equation is every solution from both roots together: x = 90°, 210° or 330°.

Check each one in $2 \sin^{2}x - \sin x - 1$. At 90°, sin x = 1, and 2 × 1 − 1 − 1 = 0. At 210° and at 330°, sin x = −½, and 2 × ¼ − (−½) − 1 = ½ + ½ − 1 = 0.

The level line at sin x = −0.5 cuts the circle at two arms, −30° and 210°, and the arm at −30° is the same position as 330°. Drag the line up to 1 and the two arms close together until they meet at the top, at 90°, where there is only one solution.

The same method with a cosine

Solve $2 \cos^{2}x + 3 \cos x - 2 = 0$ for $0^\circ \le x < 360^\circ$. Write c for cos x: $2c^{2} + 3c - 2 = 0$.

Two numbers that multiply to 2 × (−2) = −4 and add to 3 are 4 and −1. Split the middle term: $2c^{2} + 4c - c - 2 = 2c\left(c + 2\right) - 1\left(c + 2\right) = \left(2c - 1\right)\left(c + 2\right)$. So 2c − 1 = 0, giving c = ½, or c + 2 = 0, giving c = −2.

Put cos x back: cos x = ½ or cos x = −2.

A root that cannot be a cosine

The cosine of an angle is the x-coordinate of a point on a circle of radius 1, so it always lies between −1 and 1. No angle has a cosine of −2. The equation cos x = −2 has no solution, and that root is rejected. The same holds for a sine: a root of the quadratic below −1 or above 1 gives no angle.

That leaves cos x = ½. The inverse cosine gives $\cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$, and the cosine pairing gives 360° − 60° = 300°. So the solutions are x = 60° and x = 300°.

Check at 60°: cos 60° = ½, and $2 \times \frac{1}{4} + 3 \times \frac{1}{2} - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0$.

The points at 60° and 300° are the same distance across, so both angles have a cosine of ½. No point on this circle is 2 to the left of the center.

When there is no number term

Solve $2 \sin^{2}x = \sin x$ for $0^\circ \le x < 360^\circ$. It is tempting to divide both sides by sin x, which leaves 2 sin x = 1. But dividing by sin x is only allowed when sin x is not 0, and the angles where sin x is 0 are solutions of the original equation.

Instead, bring every term to one side and take out the common factor: $2 \sin^{2}x - \sin x = 0$, so sin x(2 sin x − 1) = 0. Either sin x = 0, which gives x = 0° or 180°, or sin x = ½, which gives x = 30° or 150°. There are four solutions: 0°, 30°, 150° and 180°.

The same holds for a tangent. $\tan^{2}x - \tan x = 0$ factors to tan x(tan x − 1) = 0, so tan x = 0 at 0° and 180°, and tan x = 1 at 45° and 45° + 180° = 225°.

A root that needs a calculator

Solve $3 \sin^{2}x - 2 \sin x - 1 = 0$ for $0^\circ \le x < 360^\circ$. With s = sin x it is $3s^{2} - 2s - 1 = 0$, which factors to (3s + 1)(s − 1) = 0. So $\sin x = - \frac{1}{3}$ or sin x = 1.

sin x = 1 gives x = 90°, as before. For $\sin x = - \frac{1}{3}$, the inverse sine gives $\sin^{-1}\left(- \frac{1}{3}\right) = -19.47^\circ$, to 2 decimal places. Adding a full turn gives 360° − 19.47° = 340.53°, and the sine pairing gives 180° − (−19.47°) = 199.47°.

So x = 90°, 199.47° or 340.53°. Both new angles lie between 180° and 360°, where the sine is negative, as it must be for $\sin x = - \frac{1}{3}$.

Two different ratios in one equation

Some equations mix two ratios, such as $2 \cos^{2}x + 3 \sin x - 3 = 0$. A substitution needs one ratio, so one of them must be rewritten in terms of the other.

The point at angle x on a circle of radius 1 has coordinates (cos x, sin x). The radius is the hypotenuse of a right triangle with legs cos x and sin x, so by Pythagoras $\cos^{2}x + \sin^{2}x = 1$ for every angle. That gives $\cos^{2}x = 1 - \sin^{2}x$.

Substitute: $2\left(1 - \sin^{2}x\right) + 3 \sin x - 3 = 0$, which is $-2 \sin^{2}x + 3 \sin x - 1 = 0$. Multiply every term by −1: $2 \sin^{2}x - 3 \sin x + 1 = 0$, which factors to (2 sin x − 1)(sin x − 1) = 0. So sin x = ½, giving 30° and 150°, or sin x = 1, giving 90°. The solutions are 30°, 90° and 150°.

The usual mistakes

Giving the roots of the quadratic as the answer. The roots s = −½ and s = 1 are values of sin x, not angles; each one still has to be solved for x.

Dropping a root. Each root gives its own solutions, and the answer lists all of them together. Here sin x = 1 adds 90° to the 210° and 330° from sin x = −½.

Dividing through by sin x, cos x or tan x. That loses every angle where the ratio is 0. Take out the common factor instead.

Keeping a root outside −1 to 1. No angle has a sine or cosine of −2, so cos x = −2 has no solution, and a calculator asked for $\cos^{-1}\left(-2\right)$ gives an error.