Maths Extension 1 Year 11 - Module 3 - Lesson 11
Solving Trigonometric Equations, Quadratic Type
Name
Date
Class
1. Lesson focus
Quadratic trigonometric equations. Substitute u=sinx or u=cosx, solve the quadratic, then reject roots outside [−1,1] and return to the stated interval.
Success criteria
- Apply the lesson method accurately in a routine example.
- Show each algebraic or domain step and check the result.
- Explain why the method and restrictions fit the question.
2. Worked practice
2. Solve sin²x−sinx=0 for 0≤x<2π.
3. Solve 2sin²x+sinx−1=0 for 0≤x<2π.
3. Challenge
Solve cos²x−3cosx+3=0 over the reals.
4. Multiple choice
After solving for u=sinx, which condition must a candidate root satisfy?
A. u≠0
B. u≥0
C. −1≤u≤1
D. u>1
Worked answer key
Compare each step with your working. Keep stated intervals, domains and restrictions in your final answer.
1. Solve 2cos²x−3cosx+1=0 for 0≤x<2π.
Let u=cosx: (2u−1)(u−1)=0, so cosx=1 or 1/2; x=0,π/3,5π/3.
2. Solve sin²x−sinx=0 for 0≤x<2π.
sin x(sin x−1)=0, so sin x=0 or 1. On 0≤x<2π, x=0,π,π/2; the solution set is {0,π/2,π}.
3. Solve 2sin²x+sinx−1=0 for 0≤x<2π.
(2sinx−1)(sinx+1)=0; sinx=1/2 or −1, so x=π/6,5π/6,3π/2.
Challenge. Solve cos²x−3cosx+3=0 over the reals.
Let u=cosx. Discriminant=9−12=−3<0, so there is no real cosine root and no solution.
Multiple choice. After solving for u=sinx, which condition must a candidate root satisfy?
Correct answer: C. Every real sine or cosine value lies in [−1,1]; roots outside this range cannot correspond to an angle.
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