Answer:
[tex]\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}[/tex]
Step-by-step explanation:
We are given the trigonometric equation of:
[tex]\displaystyle{\sin 4x = \dfrac{\sqrt{3}}{2}}[/tex]
Let u = 4x then:
[tex]\displaystyle{\sin u = \dfrac{\sqrt{3}}{2}}\\\\\displaystyle{\arcsin (\sin u) = \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}\\\\\displaystyle{u= \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}[/tex]
Find a measurement that makes sin(u) = √3/2 true within [0, π) which are u = 60° (π/3) and u = 120° (2π/3).
[tex]\displaystyle{u = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}[/tex]
Convert u-term back to 4x:
[tex]\displaystyle{4x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}[/tex]
Divide both sides by 4:
[tex]\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}[/tex]
The interval is given to be 0 ≤ 4x < π therefore the new interval is 0 ≤ x < π/4 and these solutions are valid since they are still in the interval.
Therefore:
[tex]\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}[/tex]
