Solve, for $0 \leq \theta < 360^{\circ}$, the equation $$9\sin(\theta + 60^{\circ}) = 4$$ giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 1

Rearranging the equation gives:

$2\tan x = 3 \sin x$

This can be manipulated as:

$\tan x = \frac{3}{2} \sin x$

Substituting $\tan x = \frac{\sin x}{\cos x}$ results in:

$\frac{\sin x}{\cos x} = \frac{3}{2} \sin x$

Assuming $\sin x \neq 0$, we can multiply both sides by $\cos x$:

$\sin x = \frac{3}{2} \sin x \cos x$

Rearranging gives:

$\sin x (1 - \frac{3}{2} \cos x) = 0$

Setting each term to zero:

1. $\sin x = 0$ gives:

   • $x = 0, -\pi, \pi$ (not included in the range)

2. Solving $1 - \frac{3}{2} \cos x = 0$,

   • Leads to $\cos x = \frac{2}{3}$.

Using $x = \cos^{-1}(\frac{2}{3})$ gives:

• $x \approx 0.8411$ and a second angle: $x \approx -0.8411$.

Thus, the solutions to two decimal places are: $x \approx 0.84, -0.84$