-Determine a general formula (or formulas) for the solutions to the following equation. Then, determine the specific solutions (if any) on the interval [0,2pi)
-Describe in the most concise way possible the general solution to the given equation, where k is any integer. Select the correct choice below and fill in all the answer boxes within your choice.
6csctheta-5=1
a) θ=__+__k
b) θ=__+__k or θ=__+__k
c) θ=__+__k or θ=__+__k or θ=__+__k
d) θ=__+__k or θ=__+__k or θ=__+__k or θ=__+__k
6csc Ø - 5 = 1
csc Ø = 6/6 = 1
then sin Ø = 1
consider the sine curve and it is clear that
Ø = π/2 or ±2π + π/2 or ±4π + π/2 ...
for the given interval, Ø = π/2
general solution:
Ø = π/2 ± 2π
no idea what all your choices mean.
Hi Reiny, those are the choices they gave me. Apparently, I have to pick the right choice and fill it out. Below is the instruction it gave me when I filled out the wrong one.
"First, isolate the trigonometric function on one side of the equation. Next determine the quadrants in which the terminal side of the argument of the function lies or determine the axis on which the terminal side of the argument of the function lies. If the terminal side of the argument lies within a quadrant, then determine the reference angle and the value(s) of the argument on the interval [0,2π). If the terminal side of the argument of the function lies along an axis, then determine the angle associated with it on the interval [0,2π). Finally, use the period of the given function to determine the general solution.
Use the most concise way to describe the general solution by making sure that the listed equations do not produce the exact same solutions. Also, ensure that none of the general solutions that have been entered can be consolidated into a fewer number of general solutions."
To find the general formula for the solutions to the equation 6csc(theta) - 5 = 1, we can start by isolating the csc(theta) term:
6csc(theta) = 6
Now, divide both sides of the equation by 6:
csc(theta) = 1
Recall that csc(theta) is equal to 1/sin(theta). So, we have:
1/sin(theta) = 1
By cross-multiplying, we get:
sin(theta) = 1
From the unit circle, we know that sin(theta) = 1 for values of theta equal to pi/2 + 2pi*k, where k is an integer. Therefore, we have:
theta = pi/2 + 2pi*k
To determine the specific solutions on the interval [0, 2pi), we substitute k values into the general formula and check if the solutions fall within the given interval.
For k = 0:
theta = pi/2 + 2pi*0 = pi/2
Since pi/2 is within the interval [0, 2pi), this is a valid solution.
For k = 1:
theta = pi/2 + 2pi*1 = pi/2 + 2pi = 5pi/2
Since 5pi/2 is outside the interval [0, 2pi), this is not a valid solution.
Therefore, the specific solution on the interval [0, 2pi) is theta = pi/2.
Based on the specific solution obtained, the most concise way to represent the general solution is:
a) θ = pi/2 + 2pi*k
To find the general formula for the solutions to the equation 6csc(theta) - 5 = 1, we start by isolating the csc(theta):
6csc(theta) = 6
Dividing both sides by 6:
csc(theta) = 1
Now, we need to determine the values of theta that satisfy this equation. Remember that csc(theta) is the reciprocal of sin(theta), so we can rewrite the equation as:
sin(theta) = 1
To find the specific solutions on the interval [0,2pi), we can use the unit circle or trigonometric identities. The sine function is equal to 1 at two points on the unit circle: pi/2 and 5pi/2.
Therefore, the specific solutions on the interval [0,2pi) are theta = pi/2 and theta = 5pi/2.
Now, let's describe the general solution in the most concise way possible. Since sin(theta) is a periodic function with a period of 2pi, we can add any integer multiple of 2pi to our specific solutions to get additional solutions. Therefore, the general solution can be represented as:
theta = pi/2 + 2pi*k
where k is any integer.
To select the correct choice, we need to fill in the blanks with the appropriate values. Since we have one specific solution (pi/2), the correct choice is:
b) theta = __ + __k or theta = __ + __k
where we fill in the blanks with pi/2, 2pi, pi/2, and 2pi.