Solve $2^x = 3$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2002 - Paper 1

To solve for $x$, we first simplify:

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

The general solution where cos is positive is:

$x = \frac{\pi}{6} + 2n\pi \text{ and } x = \frac{11\pi}{6} + 2n\pi, \text{ for } n \in \mathbb{Z}.$

By equating coefficients from the polynomial expansion, we derive:

Setting $x = -2$ shows:

a = -1, thus confirming that

a must equal $1$.

Using the identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,

we have:

$\int_{0}^{5} \sin^2 4x \, dx = \int_{0}^{5} \frac{1 - \cos(8x)}{2} \, dx$

Evaluating it from $0$ to $5$ gives:

$\left. \frac{x}{2} - \frac{\sin(8x)}{16} \right|_{0}^{5} = \frac{5}{2} - 0 = \frac{5}{2}$

Hence, the final answer is: $\frac{5}{2} \cdot 2 = 5$.

Since $AB$ is the tangent at point $A$ and $AC$ is the radius, it follows from the tangent-secant theorem that:

$\angle ACB = \angle ATB = \beta$.

From the properties of the angles we have demonstrated, it follows that:

$\angle XAY = \angle AXY$

Therefore, triangle $AXY$ is isosceles.