10. (i) Show that \( 2 \sin \left(120^{\circ}-x\right)=\cos \left(x+60^{\circ}\right) \) can be \( \operatorname{simplified} \) into \( \tan x=8-5 \sqrt{3} \). (ii) Find all the angles between \( -180^{\circ} \) and \( 180^{\circ} \) which satisfy the equation \( 2 \sin \left(120^{\circ}-\theta\right)= \) \( \cos \left(\theta+60^{\circ}\right) \).
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To simplify the equation \( 2 \sin(120^\circ - x) = \cos(x + 60^\circ) \), we start by using the sine subtraction formula: \[ \sin(120^\circ - x) = \sin 120^\circ \cos x - \cos 120^\circ \sin x. \] Since \(\sin 120^\circ = \frac{\sqrt{3}}{2}\) and \(\cos 120^\circ = -\frac{1}{2}\), we can rewrite it as: \[ \sin(120^\circ - x) = \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x. \] Multiplying by 2 gives: \[ 2\sin(120^\circ - x) = \sqrt{3} \cos x + \sin x. \] On the right side, we have: \[ \cos(x + 60^\circ) = \cos x \cos 60^\circ - \sin x \sin 60^\circ = \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x. \] Setting both sides equal: \[ \sqrt{3} \cos x + \sin x = \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x. \] Rearranging gives: \[ \left(\sqrt{3} - \frac{1}{2}\right) \cos x + \left(1 + \frac{\sqrt{3}}{2}\right) \sin x = 0. \] Simplifying, we can express this in terms of \( \tan x \): \[ \tan x = \frac{\text{(Coefficient of } \cos x\text{)}}{\text{(Coefficient of } \sin x\text{)}} = \frac{\sqrt{3}-\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} = 8 - 5\sqrt{3}. \] Now, for part (ii), we need to find all angles \( \theta \) in the range of \( -180^\circ \) to \( 180^\circ \) that satisfy: \[ 2 \sin(120^\circ - \theta) = \cos(\theta + 60^\circ). \] Since we have established that \( \tan \theta = 8 - 5 \sqrt{3} \), we find \(\theta\) by using the arctangent function: \[ \theta = \tan^{-1}(8 - 5 \sqrt{3}). \] Additionally, the tangent function is periodic with a period of \( 180^\circ \), so we can find the other solutions by adding \( 180^\circ \): \[ \theta = \tan^{-1}(8 - 5 \sqrt{3}) \quad \text{and} \quad \theta = \tan^{-1}(8 - 5 \sqrt{3}) + 180^\circ. \] However, since we are restricted to \( -180^\circ < \theta < 180^\circ \), we need to check if the second solution falls outside this range and adjust accordingly. The final solutions will be: \[ \theta \text{ in } \left( -180^\circ, 180^\circ \right). \]
