Determine the least positive value of t for which d=0.
d=cos((3pi/4)(t))
0=cos((3pi/4)(t))
I got stuck on this step
Please help
To determine the least positive value of t for which d=0, we need to solve the equation 0 = cos((3π/4)(t)).
To do that, we can start by isolating the cosine function:
cos((3π/4)(t)) = 0
Next, we can use the fact that the cosine function equals zero at certain angles. In this case, we know that cos(θ) = 0 when θ = (2n + 1)π/2 for any integer n.
So, for our equation cos((3π/4)(t)) = 0, we can set the angle inside the cosine function equal to (2n + 1)π/2 and solve for t:
(3π/4)(t) = (2n + 1)π/2
To find the least positive value of t, we want to find the smallest, positive integer value for n that satisfies the equation.
Let's solve the equation for t:
t = (2n + 1)π/2 * (4/3) (multiplied both sides by (4/3) to isolate t)
t = (2n + 1)(2π/3)
From this equation, we can see that the least positive value of t occurs when n = 0, because then t = (2(0) + 1)(2π/3) = 2π/3.
Thus, the least positive value of t for which d = 0 is t = 2π/3.