Gegee dat \(\sqrt{3}\sin x + 3 = 0\), waar \(x \in (0^{\circ}, 90^{\circ})\) - NSC Mathematics - Question 5 - 2022 - Paper 2

From the given equation (\sqrt{3}\sin x + 3 = 0), we can isolate (\sin x) as follows:

[\sin x = -\frac{3}{\sqrt{3}} = -\sqrt{3}]

Using the relationship between sine and tangent:

[\tan x = \frac{\sin x}{\cos x}]

For angles in the interval ((0^{\circ}, 90^{\circ})), (\tan x = \frac{\sin x}{\sqrt{1-\sin^2 x}}). Hence:

[\tan x = \frac{-\sqrt{3}}{\sqrt{1 - (-\sqrt{3})^2}} = \text{undefined (as the result leads to a division by zero)}]

Using the property of cosine:

[\cos(180^{\circ} + x) = -\cos x]

Thus, (\cos(180^{\circ} + x) = -\cos x).

Using the cosine and sine angle addition formulas:

[\cos(90^{\circ} + \theta) = -\sin \theta]

For sine, we have:

[\sin(\theta - 180^{\circ}) = -\sin \theta]

Substituting these results back into the expression gives us:

[\frac{-\sin \theta}{-\sin \theta + 3\sin \theta} = \frac{-\sin \theta}{2\sin \theta} = -\frac{1}{2}]

This equation can be solved by setting each factor to zero:

1. (\cos x + 2\sin x = 0) implies (\tan x = -\frac{1}{2}), which yields angles in the specified interval.
2. (3\sin 2x - 1 = 0) leads to (\sin 2x = \frac{1}{3}), subsequently solved for (x).

Starting with the left-hand side:

[LHS = \cos(x + y) \cdot \cos(x - y) = \cos^{2}x - \sin^{2}y]

This can be simplified to show:

[\cos^{2}x + \sin^{2}y = 1] confirming the identity.

Calculating gives us:

[1 - \sin 45^{\circ} - \sin 15^{\circ} = 1 - \frac{\sqrt{2}}{2} - \frac{\sqrt{6} - \sqrt{2}}{4}]

This simplifies to find the numerical value.

By factoring and simplifying:

[8\sin x \cdot \cos x = 2\sin 4x] which confirms the minimum value at (x = 67.5^{\circ}).