Given that cos(2x+15°)=sin(x−30°), 0≤x≤90°, find the value of x. In the diagram, ABCF is a straight line, BD||CE, ∠ABD=(2x+34)°, and ∠FCE=48°, find: (i) the value of x; (ii) ∠BDC

Given $\cos(2x+15^\circ) = \sin(x-30^\circ)$.
We can write $\cos(2x+15^\circ)$ as $\sin(90^\circ - (2x+15^\circ))$.
So, the equation becomes:

$$\sin(90^\circ - (2x+15^\circ)) = \sin(x-30^\circ)$$

$$90^\circ - (2x+15^\circ) = x-30^\circ$$ $$90^\circ - 2x - 15^\circ = x - 30^\circ$$

$$75^\circ - 2x = x - 30^\circ$$

Add $2x$ to both sides and add $30^\circ$ to both sides:

$$75^\circ + 30^\circ = x + 2x$$ $$105^\circ = 3x$$

Divide by 3:

$$x = \frac{105^\circ}{3}$$ $$x = 35^\circ$$

Since $0^\circ \le 35^\circ \le 90^\circ$, the value is valid.

The value of $x$ is $\boxed{\text{35}^\circ}$.