SME
Grade 12 Math · Problem type
Solve a tan x + b = c
Use this for “Solve 3 tan x + 1 = 4 for 0° ≤ x < 360°.”
Diagram
tan θ = 1 at θ = 45° and 225°, for 0° ≤ θ < 360°.
This is an example. Type your own values to start a new problem.
a
tan
x
+
b
=
c
k
k, the value of tan x
worked out (not typed)
x₁
First solution
calculated · °
x₂
Second solution
calculated · °
t₁
First solution in radians
worked out (not typed)
t₂
Second solution in radians
worked out (not typed)
Formulas
k
=
c
−
b
a
1
=
4 − 1
3
x₁
=
tan⁻¹(
k
)
45 = tan⁻¹(1)
x₂
=
x₁
+
180°
225 = 45 + 180°
t₁
=
x₁
×
π
180
π
4
=
45
×
π
180
t₂
=
x₂
×
π
180
5π/4
=
225
×
π
180
Assumptions
•
Get tan x on its own first; every value has solutions.
•
The tangent repeats every 180°, so the second solution is the first plus 180°.
•
Answers are from 0° up to 360°: add 180° to a negative tan⁻¹.
Step-by-step
WE KNOW
a = 3, b = 1, c = 4
FIND
k, the value of tan x (k), first solution (x₁), second solution (x₂), first solution in radians (t₁), second solution in radians (t₂)
Step 1 · Find k, the value of tan x (k)
Use k = (c − b) ÷ a
Take
b
from
both
sides,
then
divide
by
a
:
tan
x
on
its
own.
k
=
4 − 1
3
k
=
3
3
k
=
1
Step 2 · Find first solution (x₁)
Use x₁ = tan⁻¹(k)
tan⁻¹ gives one angle; add 180° when it is negative so it lies from 0° to 180°.
x₁
=
tan⁻¹(1)
x₁
=
45°
Step 3 · Find second solution (x₂)
Use x₂ = x₁ + 180°
Half a turn on, the opposite point has the same tangent.
x₂
=
45
+
180°
x₂
=
225°
Step 4 · Find first solution in radians (t₁)
Use t₁ = x₁ × π/180
A
half
turn,
180°,
is
π
radians:
multiply
by
π
180
.
t₁
=
45
×
π
180
t₁
=
π
4
(≈
0.79)
Step 5 · Find second solution in radians (t₂)
Use t₂ = x₂ × π/180
A
half
turn,
180°,
is
π
radians:
multiply
by
π
180
.
t₂
=
225
×
π
180
t₂
=
5π/4
(≈
3.93)
Check
Put the numbers back in. Both sides should match.
1
=
4 − 1
3
✓
tan(45°) = 1 ✓
225 = 45 + 180° ✓
π
4
=
45
×
π
180
✓
5π/4
=
225
×
π
180
✓
Topic check
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