5.3.3 Quadratic Trigonometric Equations

Quadratic trigonometric equations involve trigonometric functions that appear in a quadratic form, such as $\sin^2 \theta, \cos^2 \theta, or \tan^2 \theta.$ 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: $a \cdot (\text{trig function})^2 + b \cdot (\text{trig function}) + c = 0$ where "trig function" could be $\sin \theta,$ $\cos \theta$ , or $\tan \theta.$

2. Steps to Solve Quadratic Trigonometric Equations:

Substitute and Simplify:

If the equation is complex, it can be helpful to let u = $\text{trig function},$ where u represents $\sin \theta, \cos \theta, or \tan \theta$ , and then solve the resulting quadratic equation.

Solve the Quadratic Equation:

Factor the quadratic equation, if possible, or use the quadratic formula: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Back-Substitute:

Replace u with the original trigonometric function.

Solve the Resulting Trigonometric Equations:

Solve for $\theta$ within the specified interval.

Check the Interval:

Ensure all solutions are within the given interval, typically $0^\circ to 360^\circ \ or\ 0\ to\ 2\pi$ radians.

3. Examples of Solving Quadratic Trigonometric Equations:

infoNote

Example 1: Solving $2\sin^2 \theta - \sin \theta - 1 = 0$

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

infoNote

Example 2: Solving $\cos^2 \theta - 3\cos \theta + 2 = 0$

Step 1: Let u = $\cos \theta$ and substitute: $u^2 - 3u + 2 = 0$

Step 2: Solve the quadratic equation:
Factor the equation: $(u - 1)(u - 2) = 0$
This gives $u = 1$ or $u =$ $2$ .

Step 3: Back-substitute u = $\cos \theta:$ $\cos \theta = 1 \quad \text{or} \quad \cos \theta = 2$
Note: $\cos \theta$ = $2$ is impossible because $\cos \theta$ must be between - $1$ and $1$ .

Step 4: Solve for $\theta:$
For $\cos \theta = 1:$ $\theta = 0^\circ \quad \text{or} \quad \theta = 360^\circ$

Step 5: List all valid solutions: $\theta = :success[0^\circ, 360^\circ]$

infoNote

Example 3: Solving $\tan^2 \theta - 3 = 0$

Step 1: Let u = $\tan \theta$ and substitute: $u^2 - 3 = 0$

Step 2: Solve the quadratic equation:
Add 3 to both sides: $u^2 = 3$
Take the square root: $u = \pm \sqrt{3}$

Step 3: Back-substitute $u$ = $\tan \theta$ : $\tan \theta = \sqrt{3} \quad \text{or} \quad \tan \theta = -\sqrt{3}$
Step 4: Solve for $\theta:$
For $\tan \theta = \sqrt{3}:$ $\theta = 60^\circ \quad \text{or} \quad \theta = 240^\circ$
For $\tan \theta = -\sqrt{3}:$ $\theta = 120^\circ \quad \text{or} \quad \theta = 300^\circ$
Step 5: List all solutions: $\theta = :success[60^\circ, 120^\circ, 240^\circ, 300^\circ]$

4. Quadratic Formula for Trigonometric Equations:

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

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 $-1$ and $1$ .

Stealth Quadratics in Trigonometry

\sin(\theta) \times \sin(\theta) \equiv (\sin(\theta))^2 = \sin^2(\theta) \quad \text{(always written like this)}

\cos^2(\theta) \equiv (\cos(\theta))^2 \quad \text{and} \quad \tan^2(\theta) \equiv (\tan(\theta))^2

infoNote

Solve $2\cos^2(\theta) + 3\cos(\theta) - 1 = 0$ for $0^\circ \leq \theta \leq 360^\circ$

Let $c = \cos(\theta)$ , then:

2c^2 + 3c - 1 = 0

Use the quadratic formula:

c = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)}

c = \frac{-3 \pm \sqrt{17}}{4}

Calculating values:

c = :highlight[0.2808] \quad \text{and} \quad c = :highlight[-1.7808]

Since $-1 \leq \cos(\theta) \leq 1$ :

\cos(\theta) = 0.2808 \implies \theta = :success[73.7^\circ, 286.3^\circ] \quad \text{(to 3 s.f.)}

\cos(\theta) = -1.7808 \quad \\\text{(:highlight[Invalid solution] as it is out of range)}

Quadratic Trigonometric Equations Simplified Revision Notes for A-Level OCR Maths Pure

5.3.3 Quadratic Trigonometric Equations

1. General Form:

2. Steps to Solve Quadratic Trigonometric Equations:

3. Examples of Solving Quadratic Trigonometric Equations:

Example 1: Solving $2\sin^2 \theta - \sin \theta - 1 = 0$

Example 2: Solving $\cos^2 \theta - 3\cos \theta + 2 = 0$

Example 3: Solving $\tan^2 \theta - 3 = 0$

4. Quadratic Formula for Trigonometric Equations:

Summary:

Stealth Quadratics in Trigonometry

Solve $2\cos^2(\theta) + 3\cos(\theta) - 1 = 0$ for $0^\circ \leq \theta \leq 360^\circ$

500K+ Students Use These Powerful Tools to Master Quadratic Trigonometric Equations For their A-Level Exams.

Other Revision Notes related to Quadratic Trigonometric Equations you should explore

Quadratic Trigonometric Equations Simplified Revision Notes for A-Level OCR Maths Pure

5.3.3 Quadratic Trigonometric Equations

1. General Form:

2. Steps to Solve Quadratic Trigonometric Equations:

3. Examples of Solving Quadratic Trigonometric Equations:

Example 1: Solving 2sin⁡2θ−sin⁡θ−1=02\sin^2 \theta - \sin \theta - 1 = 02sin2θ−sinθ−1=0

Example 2: Solving cos⁡2θ−3cos⁡θ+2=0\cos^2 \theta - 3\cos \theta + 2 = 0cos2θ−3cosθ+2=0

Example 3: Solving tan⁡2θ−3=0\tan^2 \theta - 3 = 0tan2θ−3=0

4. Quadratic Formula for Trigonometric Equations:

Summary:

Stealth Quadratics in Trigonometry

Solve 2cos⁡2(θ)+3cos⁡(θ)−1=02\cos^2(\theta) + 3\cos(\theta) - 1 = 02cos2(θ)+3cos(θ)−1=0 for 0∘≤θ≤360∘0^\circ \leq \theta \leq 360^\circ0∘≤θ≤360∘

500K+ Students Use These Powerful Tools to Master Quadratic Trigonometric Equations For their A-Level Exams.

Other Revision Notes related to Quadratic Trigonometric Equations you should explore

Example 1: Solving $2\sin^2 \theta - \sin \theta - 1 = 0$

Example 2: Solving $\cos^2 \theta - 3\cos \theta + 2 = 0$

Example 3: Solving $\tan^2 \theta - 3 = 0$

Solve $2\cos^2(\theta) + 3\cos(\theta) - 1 = 0$ for $0^\circ \leq \theta \leq 360^\circ$