The trigonometric equation sin x = k asks where the rising and falling curve y = sin x meets the horizontal line at height k.
The x-coordinates of the meeting points are the solutions, and for 0 ≤ x < 2π, raising or lowering the line changes the count from two solutions to one and then to none.
An equation is solved by fixing the reference angle and picking the quadrants with the right sign; an inequality is solved by finding the boundary solutions first and then reading the interval where the curve lies above the line.
In the picture, moving the line y = k labels each meeting point with its solution, and the number of solutions appears at the top.
The meeting points are the solutions
seegongsik.com
For 0 ≤ x < 2π, the points (gold) where y = sin x meets the horizontal line y = k, with their x-coordinates
0.5
👀 Reading solutions from the graph
①The solutions of sin x = k are the x-coordinates where y = sin x meets y = k
②If −1 < k < 1 there are two solutions; if k = ±1 the line just touches a peak or trough, giving one; if |k| > 1 there are none
③When k is 0, ±1/2 or ±1 the solutions come out as neat fractions of π; otherwise the picture shows approximations
④The solution of the inequality sin x > k is the interval where the curve lies above the line
Reference angles and quadrants
Solving a trig equation
sin x = k ⇒ fix the reference angle and collect solutions in the quadrants with the right sign
For 0 ≤ x < 2π, sin x = k has two solutions when −1 < k < 1 and one when k = ±1
tan equations
tan x = k ⇒ one solution every period π
tan has period π, so for 0 ≤ x < 2π there are always two solutions
Inequalities give intervals
📐 Read whether the curve is above or below
①The solution of sin x > k is the interval of x where the curve lies above the line y = k
②First solve sin x = k to get the boundaries of the interval
③Test a point between the boundaries to see whether the curve is above or below
Worked examples
Example 1
Find all solutions of sin x = 1/2 on 0 ≤ x < 2π.
1
sin(π/6) = 1/2, so the reference angle is π/6.
2
Collect the solutions in the first and second quadrants, where sin is positive.
x = π/6, 5π/6
▸ x = π/6, 5π/6
The second-quadrant solution is π − π/6 = 5π/6. Fixing the reference angle first and picking only the quadrants with the right sign is the quick route.
Example 2
Solve cos x < 1/2 on 0 ≤ x < 2π.
1
The boundary equation cos x = 1/2 has solutions x = π/3 and 5π/3.
2
Between the two boundaries, cos x is less than 1/2.
π/3 < x < 5π/3
▸ π/3 < x < 5π/3
Testing the point x = π between them gives cos π = −1 < 1/2, so the answer is the interval between the boundaries. Find the boundaries first, then test one point in between.
Summary
Strategy
Equation: reference angle + quadrants / Inequality: boundaries, then the interval
Sketching where the solutions sit on the graph or the unit circle first is the safest approach
2022 KICE mock exam Math type, adapted
On 0 ≤ x < 2π, what is the sum of all solutions of 2 sin x = √3?
①π/2
②2π/3
③π
④4π/3
⑤3π/2
▸ ③ π
1
Dividing both sides by 2 gives sin x = √3/2, with reference angle π/3.
2
The solutions in the first and second quadrants are x = π/3 and 2π/3, so their sum is as follows.
π/3 + 2π/3 = π
🎯 Exam Points
①Find solutions as meeting points on the graph or the unit circle
②For 0 ≤ x < 2π, sin and cos equations usually have two solutions, and tan has period π
③For an inequality, find the boundary solutions first and then decide the interval between them
④Check the range of x first
⑤The special values 1/2, √2/2 and √3/2 should come to mind at once