Guides / Trigonometry

Solving trig equations: every solution, not just the calculator's

Most trig equations have two solutions in every full turn of the circle, and a calculator only ever shows you one. Nearly every lost point is a missing solution: the second quadrant nobody checked, a factor divided away, or an interval that should have been stretched.

Why it works

On the unit circle, the point at angle x is (cos x, sin x). A horizontal line at height 1/2 crosses the circle at two points, one on each side of the vertical axis, so sin x = 1/2 has two solutions per turn: π/6 and π − π/6 = 5π/6. A vertical line crosses at two points mirrored across the horizontal axis, so cosine's two solutions are θ and 2π − θ.

Which quadrants? Sine is positive in quadrants I and II, cosine in I and IV, and tangent in I and III. With a reference angle θ, the angle in quadrant II is π − θ, in III it's π + θ, and in IV it's 2π − θ. The exceptions are the values ±1 and 0, where the two points merge or sit on an axis: sin x = −1 only at 3π/2.

The calculator's inverse functions return one angle from a fixed range: arcsin and arctan give an angle between −π/2 and π/2, and arccos gives one between 0 and π. That's one solution. The other one is up to you.

Example 1: one trig function, a negative value

Solve 2 cos x + √3 = 0 on [0, 2π).

1. cos x = −√32Isolate the cosine: subtract √3, divide by 2.
2. Reference angle π/6cos(π/6) = √3/2. Use the value without its sign.
3. Quadrants II and IIICosine is negative there.
4. x = π − π/6 = 5π/6, x = π + π/6 = 7π/6One solution in each quadrant.

Example 2: a quadratic in sin x

Solve 2 sin²x + 3 sin x + 1 = 0 on [0, 2π).

1. (2 sin x + 1)(sin x + 1) = 0Factor it like 2s² + 3s + 1 = (2s + 1)(s + 1), with s = sin x.
2. sin x = −1/2 or sin x = −1Set each factor to 0, as in solving quadratics by factoring.
3. sin x = −1/2: x = 7π/6, 11π/6Reference angle π/6; sine is negative in quadrants III and IV, so π + π/6 and 2π − π/6.
4. sin x = −1: x = 3π/2The bottom of the circle. Only one point has height −1.
5. x = 7π/6, 3π/2, 11π/6Three solutions, in increasing order.

Example 3: factor, don't divide

Solve 2 sin x cos x = sin x on [0, 2π).

1. 2 sin x cos x − sin x = 0Everything on one side. Don't divide by sin x: it could be 0.
2. sin x (2 cos x − 1) = 0Factor out sin x.
3. sin x = 0: x = 0, πHeight 0 is where the circle meets the horizontal axis.
4. cos x = 1/2: x = π/3, 5π/3Reference angle π/3; cosine is positive in quadrants I and IV.
5. x = 0, π/3, π, 5π/3Four solutions. Dividing by sin x would have found only two of them (mistake 2).

Example 4: a double angle

Solve sin 2x = √32 on [0, 2π).

1. 0 ≤ 2x < 4πMultiply the interval by 2. The angle inside the sine runs through two full turns.
2. 2x = π/3, 2π/3The first turn: reference angle π/3, quadrants I and II.
3. 2x = π/3 + 2π = 7π/3, 2x = 2π/3 + 2π = 8π/3The second turn: add 2π to each.
4. x = π/6, π/3, 7π/6, 4π/3Divide every angle by 2. Two turns, two solutions each: four in all.

Example 5: a value that isn't on the unit circle chart

Solve 3 cos x + 1 = 0 on [0, 2π), to two decimal places.

1. cos x = −1/3Isolate the cosine.
2. x = arccos(−1/3) ≈ 1.91The calculator, in radian mode. arccos returns an angle between 0 and π, here in quadrant II.
3. x ≈ 1.91 or x ≈ 2π − 1.91 ≈ 4.37Cosine is also negative in quadrant III. Mirroring across the horizontal axis gives 2π − x.

Common mistakes, and the exact line they happen on

1. Stopping at the calculator's answer

Solve 3 cos x + 1 = 0 on [0, 2π) (Example 5).

1. cos x = −1/3Right so far.
2. x ≈ 1.91This is the mistake. That's the one angle arccos returns. 2π − 1.91 ≈ 4.37 has the same cosine and is also in the interval. Every value strictly between −1 and 1 has two solutions per turn.

2. Dividing by sin x

Solve 2 sin x cos x = sin x on [0, 2π) (Example 3).

1. 2 cos x = 1, so x = π/3, 5π/3This is the mistake. Dividing both sides by sin x assumes sin x isn't 0, and throws away x = 0 and x = π, where it is. Check: at x = π, both sides are 0. Factor instead, and keep sin x = 0 as one of the cases.

3. Solving 2x on only one turn

Solve sin 2x = √3/2 on [0, 2π) (Example 4).

1. 2x = π/3, 2π/3Right so far, for the first turn.
2. x = π/6, π/3This is the mistake. x runs over one turn, so 2x runs over two, and the second turn has two more solutions: 2x = 7π/3 and 8π/3, giving x = 7π/6 and 4π/3. Check: sin(2 · 7π/6) = sin(7π/3) = sin(π/3) = √3/2.

4. The wrong period for tangent

Give every solution of tan x = −1.

1. x = 3π/4Right so far: reference angle π/4, and tangent is negative in quadrant II.
2. x = 3π/4 + 2kπThis is the mistake. Tangent repeats every π, not 2π: going halfway round the circle flips the signs of both sine and cosine, and their ratio stays the same. This misses 7π/4 and every other solution in quadrant IV. The general solution is x = 3π/4 + kπ.

Practice

1. Solve 2 sin x − 1 = 0 on [0, 2π).

   Show answer

   sin x = 1/2, reference angle π/6, quadrants I and II: x = π/6, 5π/6.

2. Solve tan x = −1 on [0, 2π).

   Show answer

   Reference angle π/4; tangent is negative in quadrants II and IV: x = 3π/4, 7π/4.

3. Solve 2 cos²x − cos x = 0 on [0, 2π).

   Show answer

   Factor: cos x (2 cos x − 1) = 0. cos x = 0 gives π/2, 3π/2; cos x = 1/2 gives π/3, 5π/3. All four: x = π/3, π/2, 3π/2, 5π/3.

4. Solve cos 2x = 1/2 on [0, 2π).

   Show answer

   2x runs over [0, 4π): 2x = π/3, 5π/3, 7π/3, 11π/3. Divide by 2: x = π/6, 5π/6, 7π/6, 11π/6.

5. Solve sin x = −0.6 on [0, 2π), to two decimal places.

   Show answer

   The reference angle is arcsin 0.6 ≈ 0.64. Sine is negative in quadrants III and IV: x ≈ π + 0.64 ≈ 3.79 and x ≈ 2π − 0.64 ≈ 5.64. (The calculator's arcsin(−0.6) ≈ −0.64 isn't in the interval; adding 2π to it gives the 5.64.)