Quadratic Trigonometric Equations (AQA A-Level Mathematics): Revision Notes

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5.3.3 Quadratic Trigonometric Equations

Quadratic trigonometric equations involve trigonometric functions that appear in a quadratic form, such as sin⁡2θ,cos⁡2θ,ortan⁡2θ.\sin^2 \theta, \cos^2 \theta, or \tan^2 \theta. Solving these equations typically requires you to treat them similarly to algebraic quadratic equations, using techniques like factoring, the quadratic formula, or substitution.

1. General Form:

A quadratic trigonometric equation generally takes the form: a⋅(trig function)2+b⋅(trig function)+c=0a \cdot (\text{trig function})^2 + b \cdot (\text{trig function}) + c = 0 where "trig function" could be sin⁡θ,\sin \theta, cos⁡θ\cos \theta, or tan⁡θ.\tan \theta.

2. Steps to Solve Quadratic Trigonometric Equations:

  1. Substitute and Simplify:
  • If the equation is complex, it can be helpful to let u = trig function,\text{trig function}, where u represents sin⁡θ,cos⁡θ,ortan⁡θ\sin \theta, \cos \theta, or \tan \theta, and then solve the resulting quadratic equation.
  1. Solve the Quadratic Equation:
  • Factor the quadratic equation, if possible, or use the quadratic formula: u=−b±b2−4ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  1. Back-Substitute:
  • Replace u with the original trigonometric function.
  1. Solve the Resulting Trigonometric Equations:
  • Solve for θ\theta within the specified interval.
  1. Check the Interval:
  • Ensure all solutions are within the given interval, typically 0∘to360∘ or 0 to 2π0^\circ to 360^\circ \ or\ 0\ to\ 2\pi radians.

3. Examples of Solving Quadratic Trigonometric Equations:

infoNote

Example 1: Solving 2sin⁡2θ−sin⁡θ−1=02\sin^2 \theta - \sin \theta - 1 = 0

  • Step 1: Let u =sin⁡θ \sin \theta and substitute: 2u2−u−1=02u^2 - u - 1 = 0
  • Step 2: Solve the quadratic equation:
  • Factor the equation: (2u+1)(u−1)=0(2u + 1)(u - 1) = 0
  • This gives u =−12 or u=1. -\frac{1}{2}\ or\ u = 1.
  • Step 3: Back-substitute u = sin⁡θ:\sin \theta: sin⁡θ=−12orsin⁡θ=1\sin \theta = -\frac{1}{2} \quad \text{or} \quad \sin \theta = 1
  • Step 4: Solve for θ:\theta:
  • For sin⁡θ=−12:\sin \theta = -\frac{1}{2}: θ=210∘orθ=330∘\theta = 210^\circ \quad \text{or} \quad \theta = 330^\circ
  • For sin⁡θ=1:\sin \theta = 1: θ=90∘\theta = 90^\circ
  • Step 5: List all solutions: θ=:success[90∘,210∘,330∘]\theta = :success[90^\circ, 210^\circ, 330^\circ]
infoNote

Example 2: Solving cos⁡2θ−3cos⁡θ+2=0\cos^2 \theta - 3\cos \theta + 2 = 0

  • Step 1: Let u = cos⁡θ\cos \theta and substitute: u2−3u+2=0u^2 - 3u + 2 = 0
  • Step 2: Solve the quadratic equation:
  • Factor the equation: (u−1)(u−2)=0(u - 1)(u - 2) = 0
  • This gives u=1u = 1 or u=u = 22.
  • Step 3: Back-substitute u = cos⁡θ:\cos \theta: cos⁡θ=1orcos⁡θ=2\cos \theta = 1 \quad \text{or} \quad \cos \theta = 2
  • Note: cos⁡θ\cos \theta = 22 is impossible because cos⁡θ\cos \theta must be between -11 and 11.
  • Step 4: Solve for θ:\theta:
  • For cos⁡θ=1:\cos \theta = 1: θ=0∘orθ=360∘\theta = 0^\circ \quad \text{or} \quad \theta = 360^\circ
  • Step 5: List all valid solutions: θ=:success[0∘,360∘]\theta = :success[0^\circ, 360^\circ]
infoNote

Example 3: Solving tan⁡2θ−3=0\tan^2 \theta - 3 = 0

  • Step 1: Let u = tan⁡θ\tan \theta and substitute: u2−3=0u^2 - 3 = 0
  • Step 2: Solve the quadratic equation:
  • Add 3 to both sides: u2=3u^2 = 3
  • Take the square root: u=±3u = \pm \sqrt{3}
  • Step 3: Back-substitute uu = tan⁡θ\tan \theta: tan⁡θ=3ortan⁡θ=−3\tan \theta = \sqrt{3} \quad \text{or} \quad \tan \theta = -\sqrt{3}
  • Step 4: Solve for θ:\theta:
  • For tan⁡θ=3:\tan \theta = \sqrt{3}: θ=60∘orθ=240∘\theta = 60^\circ \quad \text{or} \quad \theta = 240^\circ
  • For tan⁡θ=−3:\tan \theta = -\sqrt{3}: θ=120∘orθ=300∘\theta = 120^\circ \quad \text{or} \quad \theta = 300^\circ
  • Step 5: List all solutions: θ=:success[60∘,120∘,240∘,300∘]\theta = :success[60^\circ, 120^\circ, 240^\circ, 300^\circ]

4. Quadratic Formula for Trigonometric Equations:

Sometimes, the quadratic cannot be factored easily, so you might need to use the quadratic formula: u=−b±b2−4ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Where u represents sin⁡θ,cos⁡θ,ortan⁡θ.\sin \theta, \cos \theta, or \tan \theta.

Summary:

Quadratic trigonometric equations are solved by first treating the trigonometric function as a variable, solving the quadratic equation, and then back-substituting the trigonometric function. Always ensure the solutions fit within the defined interval and check the validity of solutions, especially when dealing with cosine and sine, which are bounded between −1-1 and 11.

Stealth Quadratics in Trigonometry

sin⁡(θ)×sin⁡(θ)≡(sin⁡(θ))2=sin⁡2(θ)(always written like this)\sin(\theta) \times \sin(\theta) \equiv (\sin(\theta))^2 = \sin^2(\theta) \quad \text{(always written like this)} cos⁡2(θ)≡(cos⁡(θ))2andtan⁡2(θ)≡(tan⁡(θ))2\cos^2(\theta) \equiv (\cos(\theta))^2 \quad \text{and} \quad \tan^2(\theta) \equiv (\tan(\theta))^2
infoNote

Solve 2cos⁡2(θ)+3cos⁡(θ)−1=02\cos^2(\theta) + 3\cos(\theta) - 1 = 0 for 0∘≤θ≤360∘0^\circ \leq \theta \leq 360^\circ

  1. Let c=cos⁡(θ)c = \cos(\theta), then:
2c2+3c−1=02c^2 + 3c - 1 = 0
  1. Use the quadratic formula:
c=−3±32−4(2)(−1)2(2)c = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)}c=−3±174c = \frac{-3 \pm \sqrt{17}}{4}

Calculating values:

c=:highlight[0.2808]andc=:highlight[−1.7808]c = :highlight[0.2808] \quad \text{and} \quad c = :highlight[-1.7808]
  1. Since −1≤cos⁡(θ)≤1-1 \leq \cos(\theta) \leq 1:
cos⁡(θ)=0.2808  ⟹  θ=:success[73.7∘,286.3∘](to 3 s.f.)\cos(\theta) = 0.2808 \implies \theta = :success[73.7^\circ, 286.3^\circ] \quad \text{(to 3 s.f.)}imagecos⁡(θ)=−1.7808(:highlight[Invalid solution] as it is out of range)\cos(\theta) = -1.7808 \quad \\\text{(:highlight[Invalid solution] as it is out of range)}

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