IBDP Maths SL 3.6 Pythagorean identities AA HL Paper 1- Exam Style Questions- New Syllabus

Answer:
We start with \( 2\cos 2\theta – 5\cos\theta + 2 = 0 \).
Use the double-angle identity: \( \cos 2\theta = 2\cos^2\theta – 1 \).
Substitute:
\( 2(2\cos^2\theta – 1) – 5\cos\theta + 2 = 0 \)
\( 4\cos^2\theta – 2 – 5\cos\theta + 2 = 0 \)
\( 4\cos^2\theta – 5\cos\theta = 0 \)
Factor:
\( \cos\theta (4\cos\theta – 5) = 0 \)
This gives two cases:
1. \( \cos\theta = 0 \)
2. \( 4\cos\theta – 5 = 0 \) \(\Rightarrow\) \( \cos\theta = \frac{5}{4} \) (no solution since \( |\cos\theta| \le 1 \))
For \( \cos\theta = 0 \) in the interval \( \pi \le \theta \le 2\pi \):
\( \theta = \frac{3\pi}{2} \).
\(\boxed{\theta = \frac{3\pi}{2}}\)