Long Question 1 & 2

Question 1:

Solution:
(a)(b)

$\begin{array}{l}2{\text{ sin}}^{2}x=2-\frac{x}{180}\\ 1-2{\text{ sin}}^{2}x=1-\left(2-\frac{x}{180}\right)\\ \cos 2x=\frac{x}{180}-1\\ y=\frac{x}{180}-1\\ x=0,y=-1\\ x=180,y=0\\ \\ \text{Number of solutions = 2}\end{array}$

Question 2:
(a) Sketch the graph of $y=\frac{3}{2}\cos 2x\text{ for }0\le x\le \frac{3}{2}\pi .$
(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation $\frac{4}{3\pi }x-\cos 2x=\frac{3}{2}\text{ for }0\le x\le \frac{3}{2}\pi$
State the number of solutions.

Solution:
(a)(b)
$\begin{array}{l}\frac{4}{3\pi }x-\cos 2x=\frac{3}{2}\\ \cos 2x=\frac{4}{3\pi }x-\frac{3}{2}\\ \frac{3}{2}\cos 2x=\frac{3}{2}\left(\frac{4}{3\pi }x-\frac{3}{2}\right)\\ y=\frac{2}{\pi }x-\frac{9}{4}\\ \\ \text{To sketch the graph of }y=\frac{2}{\pi }x-\frac{9}{4}\\ x=0,y=-\frac{9}{4}\\ x=\frac{3\pi }{2},y=\frac{3}{4}\\ \\ \text{Number of solutions }\\ =\text{Number of intersection points}\\ \text{= 3}\end{array}$