Solutions — Trigonometric Equations with Identities
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Q1 — Quadratic in sinθ
Treat the equation as a quadratic by substituting u = sinθ:
2u² − u − 1 = 0
Factorise: (2u + 1)(u − 1) = 0
So u = −1/2 or u = 1, giving sinθ = −1/2 or sinθ = 1.
Case 1: sinθ = 1
θ = π/2 (the unique solution in [0, 2π))
Case 2: sinθ = −1/2
Reference angle: arcsin(1/2) = π/6. Sine is negative in Q3 and Q4.
Q3: θ = π + π/6 = 7π/6
Q4: θ = 2π − π/6 = 11π/6
All solutions: θ = π/2, 7π/6, 11π/6
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Q2 — Double Angle Equation I
Avoid simply writing sin 2θ = sinθ and “cancelling” — that loses solutions. Instead, move everything to one side.
Substitute sin 2θ = 2 sinθ cosθ:
2 sinθ cosθ − sinθ = 0
Factorise: sinθ(2cosθ − 1) = 0
Case 1: sinθ = 0
θ = 0 or θ = π
Case 2: cosθ = 1/2
Reference angle: π/3. Cosine is positive in Q1 and Q4.
θ = π/3 or θ = 5π/3
All solutions: θ = 0, π/3, π, 5π/3
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Q3 — Factorising a tan Equation
Do not divide by tanθ — that loses the solution tanθ = 0. Factorise:
tan²θ − tanθ = 0
tanθ(tanθ − 1) = 0
Case 1: tanθ = 0
In [0, π): θ = 0
(Note: tanθ = 0 also at θ = π, but π is not in [0, π) which is a half-open interval.)
Case 2: tanθ = 1
In [0, π): θ = π/4
All solutions: θ = 0 and θ = π/4
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Q4 — cos 2θ Equation
The equation contains cosθ, so choose the form of cos 2θ in terms of cosθ only: cos 2θ = 2cos²θ − 1.
(2cos²θ − 1) + cosθ = 0
2cos²θ + cosθ − 1 = 0
Let u = cosθ: (2u − 1)(u + 1) = 0
u = 1/2 or u = −1
Case 1: cosθ = 1/2
Reference angle π/3; Q1 and Q4: θ = π/3 or θ = 5π/3
Case 2: cosθ = −1
θ = π
All solutions: θ = π/3, π, 5π/3
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Q5 — Using Pythagorean Identity I
The equation contains sin²θ and cosθ. Replace sin²θ = 1 − cos²θ to get a single function:
3(1 − cos²θ) − 2cosθ − 2 = 0
3 − 3cos²θ − 2cosθ − 2 = 0
1 − 3cos²θ − 2cosθ = 0
Rearrange: 3cos²θ + 2cosθ − 1 = 0
Let u = cosθ: (3u − 1)(u + 1) = 0
u = 1/3 or u = −1
Case 1: cosθ = 1/3
Reference angle: α = arccos(1/3). Cosine positive in Q1 and Q4.
θ = arccos(1/3) ≈ 1.231 rad and θ = 2π − arccos(1/3) ≈ 5.052 rad
Case 2: cosθ = −1
θ = π
All solutions: θ = arccos(1/3), π, 2π − arccos(1/3)
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Q6 — Using Pythagorean Identity II
The equation contains cos²θ and sinθ. Replace cos²θ = 1 − sin²θ:
2(1 − sin²θ) + sinθ − 1 = 0
2 − 2sin²θ + sinθ − 1 = 0
1 + sinθ − 2sin²θ = 0
Rearrange: 2sin²θ − sinθ − 1 = 0
Let u = sinθ: (2u + 1)(u − 1) = 0
u = −1/2 or u = 1
Case 1: sinθ = 1 → θ = π/2
Case 2: sinθ = −1/2
Reference angle π/6; Q3 and Q4: θ = 7π/6 or θ = 11π/6
All solutions: θ = π/2, 7π/6, 11π/6
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Q7 — Double Angle Equation II
cos 2θ = cosθ − 1. The equation contains cosθ on the right, so use cos 2θ = 2cos²θ − 1:
2cos²θ − 1 = cosθ − 1
2cos²θ − cosθ = 0
cosθ(2cosθ − 1) = 0
Case 1: cosθ = 0
θ = π/2 or θ = 3π/2
Case 2: cosθ = 1/2
Reference angle π/3; Q1 and Q4: θ = π/3 or θ = 5π/3
All solutions: θ = π/3, π/2, 3π/2, 5π/3
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Q8 — General Solution
sin 2θ + √3 cos 2θ = 0
Rearrange: sin 2θ = −√3 cos 2θ
Divide both sides by cos 2θ (valid where cos 2θ ≠ 0):
tan 2θ = −√3
The reference angle is arctan(√3) = π/3. Since tan is negative in Q2 and Q4:
2θ = π − π/3 + πn = 2π/3 + πn, n ∈ ℤ
Divide by 2:
θ = π/3 + (π/2)n, n ∈ ℤ
First few solutions: θ = π/3, 5π/6, 4π/3, 11π/6, …
Note: We should verify the cases where cos 2θ = 0 (i.e. 2θ = π/2 + πk) do not satisfy the original equation: sin 2θ + √3 cos 2θ = ±1 + 0 ≠ 0, so no extra solutions arise from those cases.
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Q9 — Mixed Identity Equation
sin 2θ − √3 sinθ = 0
Substitute sin 2θ = 2 sinθ cosθ:
2 sinθ cosθ − √3 sinθ = 0
Factor out sinθ:
sinθ(2cosθ − √3) = 0
Case 1: sinθ = 0
In [0, 2π): θ = 0 or θ = π
Case 2: cosθ = √3/2
Reference angle: π/6. Cosine positive in Q1 and Q4.
θ = π/6 or θ = 2π − π/6 = 11π/6
All solutions: θ = 0, π/6, π, 11π/6
Check θ = π/6: sin(π/3) − √3 sin(π/6) = √3/2 − √3 × 1/2 = √3/2 − √3/2 = 0. ✓
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Q10 — Equation Requiring Both Identity and Factorisation
cos 2θ + 3sinθ − 1 = 0. The equation contains sinθ, so use cos 2θ = 1 − 2sin²θ:
(1 − 2sin²θ) + 3sinθ − 1 = 0
−2sin²θ + 3sinθ = 0
sinθ(3 − 2sinθ) = 0
Case 1: sinθ = 0
In [0°, 360°): θ = 0° or θ = 180°
Case 2: 3 − 2sinθ = 0 → sinθ = 3/2
This is impossible since |sinθ| ≤ 1. No solutions from this case.
All solutions: θ = 0° and θ = 180°
Check θ = 180°: cos(360°) + 3sin(180°) − 1 = 1 + 0 − 1 = 0. ✓