SME
Grade 12 Math · Problem type
Solve a sin x + b = c
Use this for “Solve 2 sin x + 3 = 4 for 0° ≤ x < 360°.”
Diagram
sin θ = 1/2 at θ = 30° and 150°, for 0° ≤ θ < 360°.
This is an example. Type your own values to start a new problem.
a
sin
x
+
b
=
c
k
k, the value of sin x
worked out (not typed)
x₁
Solution from sin⁻¹
calculated · °
x₂
Mirror solution, 180° − sin⁻¹(k)
calculated · °
t₁
x₁ in radians
worked out (not typed)
t₂
x₂ in radians
worked out (not typed)
Formulas
k
=
c
−
b
a
0.5
=
4 − 3
2
x₁
=
sin⁻¹(
k
)
30 = sin⁻¹(0.5)
x₂
=
180°
−
sin⁻¹(
k
)
150 = 180° − sin⁻¹(0.5)
t₁
=
x₁
×
π
180
π
6
=
30
×
π
180
t₂
=
x₂
×
π
180
5π/6
=
150
×
π
180
Assumptions
•
Get sin x on its own first; it must be from −1 to 1, or there is no solution.
•
The level line y = k crosses the unit circle twice: at sin⁻¹(k) and at 180° minus it (once when k = ±1).
•
Answers are from 0° up to 360°: add 360° to a negative angle.
Step-by-step
WE KNOW
a = 2, b = 3, c = 4
FIND
k, the value of sin x (k), solution from sin⁻¹ (x₁), mirror solution, 180° − sin⁻¹(k) (x₂), x₁ in radians (t₁), x₂ in radians (t₂)
Step 1 · Find k, the value of sin x (k)
Use k = (c − b) ÷ a
Take
b
from
both
sides,
then
divide
by
a
:
sin
x
on
its
own.
k
=
4 − 3
2
k
=
1
2
k
=
0.5
Step 2 · Find solution from sin⁻¹ (x₁)
Use x₁ = sin⁻¹(k)
sin⁻¹ gives one angle; add 360° when it is negative so it lies on one turn from 0°.
x₁
=
sin⁻¹(0.5)
x₁
=
30°
Step 3 · Find mirror solution, 180° − sin⁻¹(k): x₂
Use x₂ = 180° − sin⁻¹(k)
The mirror image across the y-axis has the same sine.
x₂
=
180°
−
sin⁻¹(0.5)
x₂
=
150°
Step 4 · Find x₁ in radians (t₁)
Use t₁ = x₁ × π/180
A
half
turn,
180°,
is
π
radians:
multiply
by
π
180
.
t₁
=
30
×
π
180
t₁
=
π
6
(≈
0.52)
Step 5 · Find x₂ in radians (t₂)
Use t₂ = x₂ × π/180
A
half
turn,
180°,
is
π
radians:
multiply
by
π
180
.
t₂
=
150
×
π
180
t₂
=
5π/6
(≈
2.62)
Check
Put the numbers back in. Both sides should match.
0.5
=
4 − 3
2
✓
sin(30°) = 0.5 ✓
sin(150°) = 0.5 ✓
π
6
=
30
×
π
180
✓
5π/6
=
150
×
π
180
✓
Topic check
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