Question

1. Determine the set of each of the following equations for 0° ≤ x 360° or 0° ≤ t ≤ 360°

Answer 1

Answer:

a).

$\sin(2x) = \frac{1}{2 \sqrt{2} } \\ \\ 2x = { \sin }^{ - 1} ( \frac{1}{2 \sqrt{2} } ) \\ \\ 2x = 45 \degree$

since sine is in the 2nd quadrant, other angle = 180° - x:

$2x = 45 \degree, \: 135 \degree$

we must make two rotations since it's a double angle:

$2x = 45 \degree, \: 135\degree, \: 225\degree, \: 315\degree, \: 495\degree, \: 675\degree$

divide each angle by 2:

$x = 22.5\degree, \: 67.5\degree, \: 112.5\degree, \: 157.5\degree, \: 247.5\degree, \: 337.5\degree$

Answer = {x: x = 22.5°, 67.5°, 112.5°, 157.5°, 247.5, 337.5°}

b).

$\sin(3z) = - 0.42 \\ 3z = { \sin }^{ - 1} ( - 0.42)$

since ans is negative, we take the quadrants of cosine and tangent:

$3z = 204.8\degree, \: 335.2\degree$

we make three rotations:

$3z = 384.8\degree, \: 515.2\degree, \: 695.2\degree, \: 875.2\degree, \: 1055.2\degree,$

answer is {z : z = 128.3°, 171.7°, 231.7°, 291.7°, 351.7°}