Given that $$9 \sin^2 \theta + \sin 2\theta = 8$$ show that $$8 \cot^2 \theta - 2 \cot \theta - 1 = 0$$ --- Hence, solve $$9 \sin^2 \theta + \sin 2\theta = 8$$ in the interval $$0 < \theta < 2\pi$$ Give your answers to two decimal places - AQA - A-Level Maths Pure - Question 8 - 2021 - Paper 1

Using the equation derived previously:

$9 \sin^2 \theta + \sin 2\theta = 8$

We rewrite it as:

$9 \sin^2 \theta + 2 \sin \theta \cos \theta - 8 = 0$

Let ( x = \sin \theta ), then:

$9x^2 + 2x\sqrt{1-x^2} - 8 = 0$

This equation can be solved numerically for values of ( \theta ) in the interval ( (0, 2\pi) ).

Solving numerically, we find the approximate solutions:

• ( \theta \approx 1.11 )
• ( \theta \approx 1.82 )
• ( \theta \approx 4.25 )
• ( \theta \approx 4.96 )

Hence, the answers are:

• ( \theta \approx 1.11 )
• ( \theta \approx 1.82 )
• ( \theta \approx 4.25 )
• ( \theta \approx 4.96 )

To solve the equation:

$9 \sin^2\left(2x - \frac{\pi}{4}\right) + \sin\left(4x - \frac{\pi}{2}\right) = 8$

We can express (\sin(4x - \frac{\pi}{2})) as (-\cos(4x)), leading to:

$9 \sin^2\left(2x - \frac{\pi}{4}\right) - \cos(4x) = 8$

This is a transcendental equation that can be solved graphically or numerically.

After solving, we find:

• ( x \approx 0.9 )
• ( x \approx 1.3 )

Thus, the answers are approximately:

• ( x = 0.9 )
• ( x = 1.3 )