Question 15:
Table 1 shows the values of x for the equation $$ y+3 \sin \frac{3}{2} x=1 $$

(a) Complete the values of y in Table 1.
[2 marks]
(b) For this part of the question, use the graph paper provided on page 96 . You may use a flexible curve rule.
Using a scale of 3 cm to π/3 on the x-axis and 2 cm to 1 unit on the y-axis, draw the graph of $$ y+3 \sin \frac{3}{2} x=1 \text { for } 0 \leqslant x \leqslant 2 \pi $$
[3 marks]
(c)(i) State an equation of a suitable straight line to solve $$ 3 \sin \frac{3}{2} x=\frac{4 x}{\pi}-2 . $$
(c)(ii) Hence, on the graph in (b), draw the straight line to solve $$ 3 \sin \frac{3}{2} x=\frac{4 x}{\pi}-2 . $$ State the value of x in terms of π.
[3 marks]
Solution:
(a)
$$ \begin{aligned} & y+3 \sin \frac{3}{2} x=1 \\ & y=-3 \sin \frac{3}{2} x+1 \end{aligned} $$
$$ \begin{aligned} &\text { When } x=0\\ &\begin{aligned} & y=-3 \sin \frac{3}{2}(0)+1 \\ & y=1 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & \text { When } x=\frac{\pi}{6} \\ & \begin{array}{l} \frac{\pi}{6} \times \frac{180}{\pi}=30 \\ y=-3 \sin \frac{3}{2}(30)+1 \\ y=-1.12 \end{array} \end{aligned} $$
$$ \begin{aligned} & \text { When } x=\frac{\pi}{3} \\ & \frac{\pi}{3} \times \frac{180}{\pi}=60 \\ & y=-3 \sin \frac{3}{2}(60)+1 \\ & y=-2 \end{aligned} $$

(b)
$$ \begin{aligned} y & =a \sin b x+c \\ y & =-3 \sin \frac{3}{2} x+1 \\ a & =-3, \text { it is an inverted graph } \\ c & =1 \text { (the graph moves } 1 \text { unit up) } \end{aligned} $$
$$ \begin{aligned} \text { Number of cycles } & =\frac{3}{2}=1.5 \\ \text { Period } & =2 \pi \div \frac{3}{2} \\ & =\frac{4}{3} \pi \end{aligned} $$
(c)(i)
$$ \begin{aligned} & y+3 \sin \frac{3}{2} x=1 \\ & y=1-3 \sin \frac{3}{2} x \end{aligned} $$
$$ \begin{aligned} 3 \sin \frac{3}{2} x & =\frac{4 x}{\pi}-2 \\ 0 & =\frac{4 x}{\pi}-2-3 \sin \frac{3}{2} x \\ 0 & =\frac{4 x}{\pi}-3+\left(1-3 \sin \frac{3}{2} x\right) \\ 0 & =\frac{4 x}{\pi}-3+y \\ y & =3-\frac{4 x}{\pi} \end{aligned} $$
(c)(ii)
$$ y=3-\frac{4 x}{\pi} $$
When x = 0, y = 3
When x = π, y = -1

From the graph, x coordinate of the intersection point = 114o
$$ \begin{aligned} & =114 \times \frac{\pi}{180} \\ & =\frac{19}{30} \pi \end{aligned} $$
Table 1 shows the values of x for the equation $$ y+3 \sin \frac{3}{2} x=1 $$

(a) Complete the values of y in Table 1.
[2 marks]
(b) For this part of the question, use the graph paper provided on page 96 . You may use a flexible curve rule.
Using a scale of 3 cm to π/3 on the x-axis and 2 cm to 1 unit on the y-axis, draw the graph of $$ y+3 \sin \frac{3}{2} x=1 \text { for } 0 \leqslant x \leqslant 2 \pi $$
[3 marks]
(c)(i) State an equation of a suitable straight line to solve $$ 3 \sin \frac{3}{2} x=\frac{4 x}{\pi}-2 . $$
(c)(ii) Hence, on the graph in (b), draw the straight line to solve $$ 3 \sin \frac{3}{2} x=\frac{4 x}{\pi}-2 . $$ State the value of x in terms of π.
[3 marks]
Solution:
(a)
$$ \begin{aligned} & y+3 \sin \frac{3}{2} x=1 \\ & y=-3 \sin \frac{3}{2} x+1 \end{aligned} $$
$$ \begin{aligned} &\text { When } x=0\\ &\begin{aligned} & y=-3 \sin \frac{3}{2}(0)+1 \\ & y=1 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & \text { When } x=\frac{\pi}{6} \\ & \begin{array}{l} \frac{\pi}{6} \times \frac{180}{\pi}=30 \\ y=-3 \sin \frac{3}{2}(30)+1 \\ y=-1.12 \end{array} \end{aligned} $$
$$ \begin{aligned} & \text { When } x=\frac{\pi}{3} \\ & \frac{\pi}{3} \times \frac{180}{\pi}=60 \\ & y=-3 \sin \frac{3}{2}(60)+1 \\ & y=-2 \end{aligned} $$

(b)
$$ \begin{aligned} y & =a \sin b x+c \\ y & =-3 \sin \frac{3}{2} x+1 \\ a & =-3, \text { it is an inverted graph } \\ c & =1 \text { (the graph moves } 1 \text { unit up) } \end{aligned} $$
$$ \begin{aligned} \text { Number of cycles } & =\frac{3}{2}=1.5 \\ \text { Period } & =2 \pi \div \frac{3}{2} \\ & =\frac{4}{3} \pi \end{aligned} $$

(c)(i)
$$ \begin{aligned} & y+3 \sin \frac{3}{2} x=1 \\ & y=1-3 \sin \frac{3}{2} x \end{aligned} $$
$$ \begin{aligned} 3 \sin \frac{3}{2} x & =\frac{4 x}{\pi}-2 \\ 0 & =\frac{4 x}{\pi}-2-3 \sin \frac{3}{2} x \\ 0 & =\frac{4 x}{\pi}-3+\left(1-3 \sin \frac{3}{2} x\right) \\ 0 & =\frac{4 x}{\pi}-3+y \\ y & =3-\frac{4 x}{\pi} \end{aligned} $$
(c)(ii)
$$ y=3-\frac{4 x}{\pi} $$
When x = 0, y = 3
When x = π, y = -1

From the graph, x coordinate of the intersection point = 114o
$$ \begin{aligned} & =114 \times \frac{\pi}{180} \\ & =\frac{19}{30} \pi \end{aligned} $$