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$2 sin²x − sin x − 1 = 0, to be solved for $0^{\circ }\le x<360^{\circ }$0° ≤ x < 360°. The notation ${\sin }^{2}x$sin²x means ${\left(\sin x\right)}^2$(sin x)², the sine of x, squared. It does not mean the sine of $x^2$x².

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$2s² − 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$2s² − 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)$2s² − 2s + s − 1 = 2s(s − 1) + 1(s − 1) = (2s + 1)(s − 1).

Check by expanding: $\left(2s+1\right)\left(s-1\right)=2s^2-2s+s-1=2s^2-s-1$(2s + 1)(s − 1) = 2s² − 2s + s − 1 = 2s² − 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 $y=2c^2+3c-2$y = 2c² + 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 ${\sin }^{-1}\left(-\frac{1}{2}\right)=-30^{\circ }$sin⁻¹(−½) = −30°, 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$2 sin²x − 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$2 cos²x + 3 cos x − 2 = 0 for $0^{\circ }\le x<360^{\circ }$0° ≤ x < 360°. Write c for cos x: $2c^2+3c-2=0$2c² + 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)$2c² + 4c − c − 2 = 2c(c + 2) − 1(c + 2) = (2c − 1)(c + 2). 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 ${\cos }^{-1}\left(\frac{1}{2}\right)=60^{\circ }$cos⁻¹(½) = 60°, 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$2 × ¼ + 3 × ½ − 2 = ½ + 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$2 sin²x = sin x for $0^{\circ }\le x<360^{\circ }$0° ≤ x < 360°. 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$2 sin²x − 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. ${\tan }^{2}x-\tan x=0$tan²x − 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$3 sin²x − 2 sin x − 1 = 0 for $0^{\circ }\le x<360^{\circ }$0° ≤ x < 360°. With s = sin x it is $3s^2-2s-1=0$3s² − 2s − 1 = 0, which factors to (3s + 1)(s − 1) = 0. So $\sin x=-\frac{1}{3}$sin x = −1/3 or sin x = 1.

sin x = 1 gives x = 90°, as before. For $\sin x=-\frac{1}{3}$sin x = −1/3, the inverse sine gives ${\sin }^{-1}\left(-\frac{1}{3}\right)=-19.47^{\circ }$sin⁻¹(−1/3) = −19.47°, 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 $\sin x=-\frac{1}{3}$sin x = −1/3.

Two different ratios in one equation

Some equations mix two ratios, such as $2{\cos }^{2}x+3\sin x-3=0$2 cos²x + 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 ${\cos }^{2}x+{\sin }^{2}x=1$cos²x + sin²x = 1 for every angle. That gives ${\cos }^{2}x=1-{\sin }^{2}x$cos²x = 1 − sin²x.

Substitute: $2\left(1-{\sin }^{2}x\right)+3\sin x-3=0$2(1 − sin²x) + 3 sin x − 3 = 0, which is $-2{\sin }^{2}x+3\sin x-1=0$−2 sin²x + 3 sin x − 1 = 0. Multiply every term by −1: $2{\sin }^{2}x-3\sin x+1=0$2 sin²x − 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 ${\cos }^{-1}\left(-2\right)$cos⁻¹(−2) gives an error.