Long Question 6

Question 6 (10 marks):
(a) Prove that 2 tan x cos² x = sin 2x.

(b) Hence, solve the equation 4 tan x cos² x = 1 for 0 ≤ x ≤ 2π.

(c)(i) Sketch the graph of y = sin 2x for 0 ≤ x ≤ 2π.

(c)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions for the equation 4π tan x cos² x = x – 2π for 0 ≤ x ≤ 2π.
State the number of solutions.

Solution: 
(a)
$\begin{array}{l}2\tan x{\cos }^{2}x=\sin 2x\\ \text{Left hand side}\\ =2\tan x{\cos }^{2}x\\ =2\times \frac{\sin x}{\cos x}\times {\cos }^{2}x\\ =2\sin x\cos x\\ =\sin 2x\\ =\text{ Right hand side }\left(\text{Proven}\right)\end{array}$


(b)
$\begin{array}{l}4\tan x{\cos }^{2}x=1,0\le x\le 2\pi \\ 2\left(2\tan x{\cos }^{2}x\right)=1\\ 2\sin 2x=1\\ \sin 2x=\frac{1}{2}\\ \text{Basic angle}=\frac{\pi }{6}\\ 2x=\frac{\pi }{6},\left(\pi -\frac{\pi }{6}\right),\left(2\pi +\frac{\pi }{6}\right),\left(3\pi -\frac{\pi }{6}\right)\\ 2x=\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}\\ x=\frac{\pi }{12},\frac{5\pi }{12},\frac{13\pi }{12},\frac{17\pi }{12}\end{array}$



(c)(i)
y = sin 2x, 0 ≤ x ≤ 2π.




(c)(ii)
$\begin{array}{l}4\pi \tan x{\cos }^{2}x=x-2\pi \\ 2\pi \left(2\tan x{\cos }^{2}x\right)=x-2\pi \\ 2\pi \sin 2x=x-2\pi \\ \sin 2x=\frac{x}{2\pi }-\frac{2\pi }{2\pi }\\ \sin 2x=\frac{x}{2\pi }-1\\ y=\frac{x}{2\pi }-1\end{array}$


Number of solutions = 4