Before or after solving a trig equation, sketch the graph of the trig function over the stated interval, draw the horizontal line for the right-hand side, and count the crossings. That count is how many solutions you should have. This appears in every “find all solutions” question, and it is the fastest self-check in the topic.
The skill works alongside the basic angle and quadrant routine and the widened interval after a substitution. Both can produce a list, but only the sketch tells you whether the list is complete.
How do you count solutions from a sketch?
- Find the period of the function, for example 360°/2 = 180° for sin 2x, using what you learned about period.
- Count the cycles inside the interval by dividing the interval length by the period.
- Draw the line for the right-hand side. A value strictly between −1 and 1 crosses sine or cosine twice per cycle. Tangent crosses any value once per 180° cycle.
- Write the expected count beside the question, then compare it with your list.
If the numbers do not match, a solution is missing or wrong. Look for the cycle with no crossing marked.
Worked example
Solve sin 2x = 0.6 for 0° ≤ x ≤ 360°.
Step 1, sketch first: the period of sin 2x is 180°, so there are two cycles in 360°. The line y = 0.6 crosses each cycle twice. Expect 4 solutions.
Step 2, solve for u = 2x: the interval for u is 0° ≤ u ≤ 720°. The basic angle is sin⁻¹(0.6) = 36.87°. In the first cycle, u = 36.87° and 180° − 36.87° = 143.13°. In the second, u = 36.87° + 360° = 396.87° and 143.13° + 360° = 503.13°.
Step 3, convert: x = u/2, so x = 18.43°, 71.57°, 198.43°, 251.57°.
Answer: x = 18.4°, 71.6°, 198.4°, 251.6° (to 1 d.p.).
Check: the list has 4 values, matching the sketch. Substituting, sin(2 × 71.57°) = sin 143.13° = 0.6.
The mistake to watch for
A common slip is to stop after finding two answers, because two is what the basic method gives.
Mistaken answer: x = 18.4° and 71.6°
The student solved sin u = 0.6 once and divided by 2, but never asked how many cycles there were. Only the first half of the interval is covered.
The correction is to draw the two waves before calculating. The sketch shows two humps, each crossed twice by the line, and the empty second hump points at the missing solutions. After a sketch, the list of four values is easy to verify against the picture.
Check yourself
1. How many solutions does cos 3x = 0.2 have for 0° ≤ x ≤ 360°?
Show answer
The period is 360°/3 = 120°, so there are 3 cycles in 360°. The value 0.2 lies between −1 and 1, so each cycle gives two crossings: 3 × 2 = 6.
6 solutions
2. Solve tan 2x = 1 for 0° ≤ x ≤ 180°. How many solutions does your sketch predict?
Show answer
The period of tan 2x is 90°, so there are 2 cycles in 180°, and tangent takes each value once per cycle: 2 solutions. Let u = 2x, 0° ≤ u ≤ 360°. tan u = 1 at u = 45° and 225°, so x = 22.5° and 112.5°.
x = 22.5°, 112.5°
3. Solve sin(x/2) = 0.8 for 0° ≤ x ≤ 720°.
Show answer
The period is 720°, so one cycle: expect 2 solutions. Let u = x/2, 0° ≤ u ≤ 360°. sin⁻¹(0.8) = 53.13°, so u = 53.13° and 126.87°. Then x = 2u.
x = 106.3° and 253.7°
Where this leads next
The remaining lesson, interpret an amplitude change separately from translation, shows what a sketch must include when the curve is also stretched and lifted. Then test everything in the mixed practice set. The quadratic structure explorer builds the same habit of checking a graph against the numbers, and the non-calculator working trainer sharpens exact values.
If you often miss answers even with a sketch, a teacher can check your graphs step by step in online one-to-one Additional Mathematics tuition.