sin(3x + pi/18) = 1 Solve for interval 0 less theta less 2pi

Answer 1

Then 3x+pi/18 = pi/2 must hold because that is the only place where the sine is one.

3x = pi/2 - pi/18 = (9/18) pi - (1/18) pi = (8/18) pi

x = (8/54) pi = (4/27)pi

Answer 2

Sine function has value of 1 at θ≤ =π/2 between 0≤θ≤2π

so 3x +π/18 = π/2

3x = π/2 - π/18

3x = 8π/18 =4π/9

x = 4π/27