That's better. For f(x) = 0, you need
sin(2x-60) = 1/3
Let sinθ = 1/3, so θ = 19.4°
2x - 60 = 19.4°
2x = 60 + 19.4°
x = 30 + 9.7°
Figuring that sin is positive in 1st and 2nd quadrants,
x = 180n + 120 - 9.7 = 180n + 110.3
for any integer n
consider the function:
f(x)=3sin(2(x-30))-1
determine the first three x-intercepts to the right of the origin, rounded to the nearest tenth of a degree.
1 answer
1 answer