SME
Grade 12 Math · Problem type
A quadratic in sin x
Use this for “Solve 2 sin²x − sin x − 1 = 0 for 0° ≤ x < 360°.”
Diagram
sin θ = −1/2 at θ = 210° and 330°
sin θ = 1 at θ = 90°
3 solutions for 0° ≤ θ < 360°
This is an example. Type your own values to start a new problem.
a
sin²x
+
(
b
)
sin
x
+
(
c
)
=
0
s₁
Smaller value of sin x
worked out (not typed)
s₂
Larger value of sin x
worked out (not typed)
Formulas
s₁
=
−
b
−
√
b
2
− 4 ×
a
×
c
2 ×
a
−
1
2
=
(−(−1)
−
√(
(−1)
2
−
4
×
2
×
(−1)))
÷
(2
×
2)
s₂
=
−
b
+
√
b
2
− 4 ×
a
×
c
2 ×
a
1
=
(−(−1)
+
√(
(−1)
2
−
4
×
2
×
(−1)))
÷
(2
×
2)
Assumptions
•
Treat sin x as one unknown s: 2 sin²x − sin x − 1 = 0 is the quadratic 2s² − s − 1 = 0.
•
Its two roots are two values of sin x, and each gives its own angles.
•
A root less than −1 or more than 1 gives no angle, since sin x stays from −1 to 1.
Step-by-step
WE KNOW
a = 2, b = −1, c = −1
FIND
smaller value of sin x (s₁), larger value of sin x (s₂)
Step 1 · Find smaller value of sin x (s₁)
Use s₁ = (−b − √(b² − 4 × a × c)) ÷ (2 × a)
The quadratic formula in s = sin x: the smaller root.
s₁
=
(−(−1)
−
√(
(−1)
2
−
4
×
2
×
(−1)))
÷
(2
×
2)
s₁
=
1 −
√
9
4
s₁
=
1 − 3
4
s₁
=
−
1
2
→
x
=
210°
or
330°
Step 2 · Find larger value of sin x (s₂)
Use s₂ = (−b + √(b² − 4 × a × c)) ÷ (2 × a)
And
the
larger
root;
when
b
2
−
4ac
=
0
the
two
are
the
same.
s₂
=
(−(−1)
+
√(
(−1)
2
−
4
×
2
×
(−1)))
÷
(2
×
2)
s₂
=
1 +
√
9
4
s₂
=
1 + 3
4
s₂
=
1
→
x
=
90°
→
all
solutions:
x
=
90°,
210°
or
330°
Check
Put the numbers back in. Both sides should match.
−
1
2
=
(−(−1)
−
√(
(−1)
2
−
4
×
2
×
(−1)))
÷
(2
×
2)
✓
1
=
(−(−1)
+
√(
(−1)
2
−
4
×
2
×
(−1)))
÷
(2
×
2)
✓
Topic check
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