Given that g(x)=5sin2x+12cos2x, express g(x) in the form Rain(2x+alpha), where R>0 and alpha is an acute angle. Find the minimum value of 26/g(x)+15

We are given $g(x) = 5\sin(2x) + 12\cos(2x)$.
We want to express this in the form $R\sin(2x+\alpha)$, where $R > 0$ and $\alpha$ is an acute angle.
Using the compound angle formula, $R\sin(2x+\alpha) = R(\sin(2x)\cos\alpha + \cos(2x)\sin\alpha)$.

$$R\sin(2x+\alpha) = (R\cos\alpha)\sin(2x) + (R\sin\alpha)\cos(2x)$$

Comparing this to $5\sin(2x) + 12\cos(2x)$, we have:

$$R\cos\alpha = 5 \quad \text{(1)}$$ $$R\sin\alpha = 12 \quad \text{(2)}$$

To find $R$, square and add (1) and (2):

$$(R\cos\alpha)^2 + (R\sin\alpha)^2 = 5^2 + 12^2$$ $$R^2(\cos^2\alpha + \sin^2\alpha) = 25 + 144$$

Since $\cos^2\alpha + \sin^2\alpha = 1$:

$$R^2 = 169$$

Since $R > 0$:

$$R = \sqrt{169} = 13$$

To find $\alpha$, divide (2) by (1):

$$\frac{R\sin\alpha}{R\cos\alpha} = \frac{12}{5}$$ $$\tan\alpha = \frac{12}{5}$$

Since $\alpha$ is an acute angle, we find its value:

$$\alpha = \arctan\left(\frac{12}{5}\right) \approx 67.38^\circ$$

Therefore, $g(x)$ can be expressed as:

$$g(x) = 13\sin(2x + \arctan\left(\frac{12}{5}\right))$$

or approximately

$$g(x) = 13\sin(2x + 67.38^\circ)$$

Substitute the expression for $g(x)$ into the given expression:

$$\frac{26}{g(x)} + 15 = \frac{26}{13\sin(2x+\alpha)} + 15$$ $$= \frac{2}{\sin(2x+\alpha)} + 15$$

The range of the sine function is $-1 \le \sin(2x+\alpha) \le 1$.
For the expression $\frac{2}{\sin(2x+\alpha)} + 15$ to be minimum, the term $\frac{2}{\sin(2x+\alpha)}$ must be minimum.
This occurs when $\sin(2x+\alpha)$ is negative and has the largest possible magnitude, which is $\sin(2x+\alpha) = -1$.
Substitute $\sin(2x+\alpha) = -1$ into the expression:

$$\text{Minimum value} = \frac{2}{-1} + 15$$ $$= -2 + 15$$ $$= 13$$

The minimum value of $\frac{26}{g(x)} + 15$ is $\boxed{13}$.