8 (a) Given that 9 sin² θ + sin 2θ = 8 show that 8 cot² θ - 2 cot θ - 1 = 0 8 (b) Hence, solve 9 sin² θ + sin 2θ = 8 in the interval 0 < θ < 2π - AQA - A-Level Maths Mechanics - Question 8 - 2021 - Paper 1

To show that

8 cot² θ - 2 cot θ - 1 = 0,

we start with the equation:

1. Recall that sin 2θ = 2 sin θ cos θ.

2. Substitute sin 2θ in the equation to get:

   9 sin² θ + 2 sin θ cos θ = 8.

3. Rearranging gives:

   9 sin² θ + 2 sin θ cos θ - 8 = 0.

4. Divide by sin² θ (assuming sin θ ≠ 0):

   9 + 2 rac{ ext{cos} θ}{ ext{sin} θ} - 8 rac{1}{ ext{sin}² θ} = 0.

5. Let

   u = rac{ ext{cos} θ}{ ext{sin} θ} = ext{cot} θ.

6. Hence, the equation can be rewritten as:

   8 ext{cot}² θ - 2 ext{cot} θ - 1 = 0.