Maths help

This is trigonometry.

4(a)

$\text{As }\sin x = \cos\left(\dfrac{\pi}{2}-x\right),\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{2}-\dfrac{\pi}3\right)=\cos\left(\pi\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\right)=\cos\left(\pi\left(\dfrac{1}{6}\right)\right) =\cos\left(\dfrac{\pi}{6}\right)\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \cos\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \text{Comparing the left hand side and the right hand side of the equation,}\\ k =6$

4(b)

$\cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right)\\ \text{As the cosine function is periodic with period }2\pi\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}+2n\pi\right)\\ \text{As }\cos x = \cos\left(-x\right)\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(-\dfrac{\pi}{6}+2n\pi\right)\\ \text{Comparing the left hand side and the right hand side of the equation, }\\ 2x-\dfrac{5\pi}{6} = \dfrac{\pi}{6} + 2n\pi \text{ or } 2x-\dfrac{5\pi}{6} = -\dfrac{\pi}{6} + 2n\pi\\ 2x = \pi+2n\pi\text{ or }2x=\dfrac{2\pi}{3}+2n\pi\\ x = \dfrac{\pi}{2}+n\pi\text{ or }x=\dfrac{\pi}{3}+n\pi$

4(c)

$\text{As shown above in 4(b), } \cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\implies x = n\pi+\dfrac{\pi}{2}\text{ or }x=n\pi+\dfrac{\pi}{3}\\ \text{As tan} x \text{ has a period of }\pi,\\ \tan x = \tan\left(\dfrac{\pi}{2}\right)\text{ or }\tan x = \tan\left(\dfrac{\pi}{3}\right)\\ \text{But considering the fact that}\tan x \text{ has an asymptote at } x = \dfrac{\pi}{2},\text{ the only finite value is }\tan\left(\dfrac{\pi}{3}\right).\\ \text{The required value is }\tan\left(\dfrac{\pi}{3}\right) = \sqrt3$