Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.)
cos θ = −1/2
1).θ= Answer-->(2π/3)+2πn, 4π/3+2πn
2). List six specific solutions
a).
b).
c).
d).
e).
f).
I really need help getting these please. I've tried 3 times already.
Oh, I see you're cosplaying as someone who needs help with trigonometry! Don't worry, I've got your back! Let's solve this equation and find those solutions.
Since you were given cos θ = -1/2, we know that θ must be in either the second or third quadrant, where the cosine value is negative. In these quadrants, the reference angle θ' can be found through the inverse cosine of 1/2, which is π/3.
Now let's find the general solutions:
1) In the second quadrant, θ = π - θ', which gives us θ = π - π/3 = 2π/3.
2) In the third quadrant, θ = π + θ', which gives us θ = π + π/3 = 4π/3.
Excellent! Now add 2πn, where n is any integer, to both solutions to find more specific solutions.
a) θ = 2π/3 + 2π(0) -> 2π/3
b) θ = 4π/3 + 2π(0) -> 4π/3
c) θ = 2π/3 + 2π(1) -> 8π/3
d) θ = 4π/3 + 2π(1) -> 10π/3
e) θ = 2π/3 + 2π(-1) -> -4π/3
f) θ = 4π/3 + 2π(-1) -> -2π/3
There you have it! Six specific solutions for the equation cos θ = -1/2. Feel free to round those angles to two decimal places if you'd like. Have fun with your math adventures!
To solve the equation cos θ = -1/2, we need to find the values of θ that satisfy this equation.
1) To find the general solutions, we can use the inverse cosine function or the unit circle. Since the cosine function has a period of 2π, we can add or subtract any integer multiple of 2π to the solutions.
The values of θ that satisfy the equation cos θ = -1/2 are given by:
θ = (2π/3) + 2πn and θ = (4π/3) + 2πn, where n is an integer.
So the general solutions are θ = (2π/3) + 2πn and θ = (4π/3) + 2πn.
2) To list six specific solutions, we can substitute different values for n. Let's start with n = 0 and increment it by 1 each time:
a) θ = (2π/3) + 2π(0) = 2π/3
b) θ = (4π/3) + 2π(0) = 4π/3
c) θ = (2π/3) + 2π(1) = 8π/3
d) θ = (4π/3) + 2π(1) = 10π/3
e) θ = (2π/3) + 2π(2) = 14π/3
f) θ = (4π/3) + 2π(2) = 16π/3
So, six specific solutions are:
a) θ = 2π/3
b) θ = 4π/3
c) θ = 8π/3
d) θ = 10π/3
e) θ = 14π/3
f) θ = 16π/3
To solve the equation cos θ = -1/2, we need to find values of θ that satisfy this equation.
First, we can look at the unit circle. Recall that the unit circle is a circle with a radius of 1 centered at the origin on a coordinate plane. The cosine of an angle θ on the unit circle gives the x-coordinate of the point where the terminal side of the angle intersects the unit circle. In our case, we are looking for points where the x-coordinate is -1/2.
Looking at the unit circle, we see that there are two points where the x-coordinate is -1/2, corresponding to angles of approximately 2π/3 and 4π/3.
To find the other values of θ, we can use the periodicity of the cosine function. The cosine function has a period of 2π, which means that if one value of θ satisfies the equation, all other values can be found by adding or subtracting integer multiples of 2π.
So, our general solution for θ is θ = 2π/3 + 2πn and θ = 4π/3 + 2πn, where n is any integer.
To find specific solutions, we can substitute different values for n and find the corresponding θ values. Let's find six specific solutions:
a) When n = 0:
θ = 2π/3 + 2π(0)
= 2π/3
b) When n = 1:
θ = 2π/3 + 2π(1)
= 2π/3 + 2π
= 8π/3
c) When n = 2:
θ = 2π/3 + 2π(2)
= 2π/3 + 4π
= 14π/3
d) When n = -1:
θ = 2π/3 + 2π(-1)
= 2π/3 - 2π
= -4π/3
e) When n = -2:
θ = 2π/3 + 2π(-2)
= 2π/3 - 4π
= -10π/3
f) When n = -3:
θ = 2π/3 + 2π(-3)
= 2π/3 - 6π
= -16π/3
So, the six specific solutions are:
a) θ = 2π/3
b) θ = 8π/3
c) θ = 14π/3
d) θ = -4π/3
e) θ = -10π/3
f) θ = -16π/3
Remember to round terms to two decimal places as requested.
you know that the reference angle is θ = π/3, right? Because cos π/3 = 1/2
Now, since cosπ = x/r, you need x negative, so you are in QII and QIII.
So, draw your triangle on the negative x-axis, giving you π-π/3 and π+π/3 = 2π/3 and 4π/3.
Since cosine has a period of 2π, you can add or subtract 2π as many times as you like and get the same value for cosθ
That gives you the answer in part 1.
Now, having those solutions, just plug in different values of n:
n = 1: 2π/3 + 2π*1 = 8π/3
n = -3: 4π/3 + 2π(-3) = -14π/3
and so on for as many values as you need.