1
Two quadrants per value, always.
sin x positive
quadrants I and II
π − reference
cos x positive
quadrants I and IV
2π − reference
tan x positive
quadrants I and III
π + reference
A calculator returns the principal value only — one angle out of two in [0, 2π), and
out of infinitely many overall. Finding the second one is your job, not the
calculator's. That is what makes this an equation, not an identity: an identity
such as sin²x + cos²x = 1 holds for every angle, while an equation holds only for
particular angles, and the work is finding them all.
2
Reference angle, then both quadrants.
reference angle: cos⁻¹(√32) = π6
ignore the sign for now
cosine negative in II and III
where x is negative
II: π − π6 = 5π6
first solution
III: π + π6 = 7π6
second solution
Work with the size first and apply the sign afterwards. Feeding the negative straight
into a calculator gives an angle outside [0, 2π), which then has to be converted anyway.
3
Factor — never divide by a trigonometric factor.
2cos²x − cos x = 0
common factor cos x
cos x(2 cos x − 1) = 0
factor, do not divide
cos x = 0 → x = π2, 3π2
from the first factor
cos x = 12 → x = π3, 5π3
four solutions in all
Dividing by cos x throws away two answers
Cancelling cos x from both sides to get 2 cos x = 1 quietly assumes cos x ≠ 0, and
loses x = π2 and x = 3π2. Exactly the same trap as dividing a quadratic by x. Factor
and set each piece to zero.
4
Use an identity to get down to one function.
2sin²x + 3cos x − 3 = 0
two functions, cannot solve yet
sin²x = 1 − cos²x
the Pythagorean identity
2 − 2cos²x + 3cos x − 3 = 0
now only cosine
2cos²x − 3cos x + 1 = 0
a quadratic in cos x (times −1)
It factors as (2 cos x − 1)(cos x − 1) = 0, giving cos x = 12 or cos x = 1, and so
x = π3, 5π3 and 0.
5
Every solution, not just one turn: add whole periods.
An interval like [0, 2π) asks for one turn. "All solutions" asks for every turn, written once with an integer n.
Write every solution of cos x = √22.
one turn: x = π4 and 7π4
quadrants I and IV
x = π4 + 2πn or x = 7π4 + 2πn
n any integer: add whole turns
Each answer from one turn repeats every 2π, so each gets its own "+ 2πn".
Write every solution of tan x = 1.
one turn: x = π4 and 5π4
quadrants I and III
5π4 = π4 + π
the second is the first plus half a turn
x = π4 + πn
tangent repeats every π
Tangent's period is π, so one family "+ πn" covers both answers.
6
Variations: no solution, one solution, and a tangent equation.
Isolate the function first; the number it equals decides how many answers there are.
No solution: 2 cos x + 3 = 0
−32 is below −1
cosine never leaves [−1, 1]
no solution
a complete answer
Check the size before looking for a reference angle. Anything outside [−1, 1] has no sine or cosine angle.
Only one solution: 4 sin x + 7 = 3
sin x = −1
isolate: 4 sin x = −4
the lowest point of the unit circle
an axis point, not a quadrant
x = 3π2 only
no second quadrant to pair with
At ±1 and 0 the "two quadrants" rule does not apply; read the angle straight off the axes.
A tangent equation: √3 tan x − 1 = 0
tan x = 1√3 = √33
isolate
reference angle π6
tan 30° = √33
tangent positive in I and III
π6 and π + π6
x = π6, 7π6
half a turn apart
Tangent's two answers are always exactly π apart, unlike sine's and cosine's.
7
A multiple angle means widening the search first.
x in [0, 2π) means 2x in [0, 4π)
double the interval too
2x = π3, 5π3, 7π3, 11π3
two turns, so four answers
x = π6, 5π6, 7π6, 11π6
halve every one
x = π6 and 5π6
Solving for 2x in the original interval and halving at the end finds only two of the
four. Widen the interval to match the multiple angle first — sin 3x over [0, 2π)
needs 3x searched over [0, 6π), and yields six solutions.
Remember the flow
Isolate the function · Reference angle from the size · Both quadrants for the sign · Factor rather than divide · Widen the interval for multiple angles
A calculator gives one answer. The interval usually contains two or more.
On the exam
Tests ask for all solutions in [0, 2π) or 0° to 360°, and the trap choices give only the calculator's one angle, or lose answers by dividing. The fastest route: isolate the function, find the reference angle from its size, place it in both quadrants the sign allows, and widen the interval for a multiple angle. A 10-second check: substitute each answer back into the original equation.
