find all the values of X between 0 and 2π such that cos(3x-π/2) = √3/2
cos theta = sqrt ( 3 ) / 2
for
theta = pi /6
and
and theta = 11 pi / 6
Solutions :
3 x - pi / 2 = pi / 6
3 x = pi / 6 + pi / 2
3 x = pi / 6 + 3 pi / 6
3 x = 4 pi / 6
3 x = 2 pi / 3 Divide both sides by 3
x = 2 pi / 9
AND
3 x - pi / 2 = 11 pi / 6
3 x = 11 pi / 6 + pi / 2
3 x = 11 pi / 6 + 3 pi / 6
3 x = 14 pi / 6
3 x = 7 pi / 3 Divide both sides by 3
x = 7 pi / 9
Remark:
pi / 6 = 30 °
11 pi / 6 = 330 °
2 pi / 9 = 40 °
7 pi / 9 = 140 °