What are the solutions of the equation 2sin(4x)=1 on the interval [0,π)?

Answer 1

2sin(4x) = 1

Divide each term by 2 

Sin(4x) = ½

Take the inverse sine (arcsine) of both sides to extract 4x from inside the sine

4x = arcsine (1/2)

The exact value of arcsine (1/2) is pi/6

4x = pi/6

4x = pi/6 + 2(pi)n, n = 0,1, 2,3,…..

Divide each term by 4 and simplify.

x = pi/24 + (1/2)(pi)n, n = 0,1, 2,3,…..

The final solution is

 x = pi/24 + (1/2)(pi)n, for any integer n

Answer 2

Let u=4x

2 sin(4x) = 1 Substitute in u

2 sin(u) = 1 Divide both sides by 2

2 sin(u)/2 = 1/2

sin(u)= 1/2

Using the unit circle, the first quadrant angle with a sin = 1/2 is u=π/6 In the second quadrant, sin is also positive. So the other solution in the second quadrant with the same reference angle is u=5π/6

Now substituting back into u

u=4x

π/6=4x Divide both sides by 4

x=π/24

5π/6=4x Divide both sides by 4

x=5π/24