Question

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Answer 1

Answer:

$\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}$

Step-by-step explanation:

We are given the trigonometric equation of:

$\displaystyle{\sin 4x = \dfrac{\sqrt{3}}{2}}$

Let u = 4x then:

$\displaystyle{\sin u = \dfrac{\sqrt{3}}{2}}\\\\\displaystyle{\arcsin (\sin u) = \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}\\\\\displaystyle{u= \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}$

Find a measurement that makes sin(u) = √3/2 true within [0, π) which are u = 60° (π/3) and u = 120° (2π/3).

$\displaystyle{u = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}$

Convert u-term back to 4x:

$\displaystyle{4x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}$

Divide both sides by 4:

$\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}$

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:

$\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}$