sin(2θ)=1/2
Find all solutions in the interval [0, 2π). (Enter your answers as a comma-separated list.) k = any integer
What I have so far: pi/12 + kpi, 5pi/12 + kpi
However, I do not know how to solve for the interval. Please help, thank you!
θ = π / 12 + k π and θ = 5 π / 12 + k π
just put k = 0 , ±1 , ±2 , ±3... in this solutions
k = - 1
θ = π / 12 + k π = π / 12 - 1 ∙ π = π / 12 - π = π / 12 - 12 π / 12 = - 11 π / 12
θ = π / 12 + k π = 5 π / 12 - 1 ∙ π = 5 π / 12 - π = 5 π / 12 - 12 π / 12 = - 7 π / 12
θ = - 11 π / 12 and θ = - 7 π / 12 are less than 0, lie outside the interval [ 0 , 2 π ) and they are not solutions.
For k = - 2 , - 3 , - 4 ... θ is also less than 0, lie outside the interval and they are not solutions.
k = 0
θ = π / 12 + k π = π / 12 + 0 ∙ π = π / 12
θ = π / 12 + k π = 5 π / 12 + 0 ∙ π = 5 π / 12
k = 1
θ = π / 12 + k π = π / 12 + 1 ∙ π = π / 12 + π = π / 12 + 12 π / 12 = 13 π / 12
θ = π / 12 + k π = 5 π / 12 + 1 ∙ π = 5 π / 12 + π = 5 π / 12 + 12 π / 12 = 17 π / 12
k = 2
θ = π / 12 + k π = π / 12 + 2 π
θ = π / 12 + k π = 5 π / 12 + 2 π
θ = π / 12 + 2 π and θ = 5 π / 12 + 2 π are greater than 2 π, lie outside the interval [ 0 , 2 π ) and they are not solutions.
For k = 3 , 4 , 5 ... θ is also greater than 2 π, lie outside the interval and they are not solutions.
So the solutons are:
π / 12 , 5 π / 12 , 13 π / 12 , 17 π / 12
so, you have your reference angle π/6
Now, sin2θ has period 2π/2 = π, so all solutions are the values
π/6 + nπ: π/6, 7π/6
5π/6 + nπ: 5π/6, 11π/6
where you just include the ones in [0,2π)
кАЩх
Why did the angle go to the party without its interval buddy? Because it wanted to have a sinfully good time! But don't worry, I'll help you find the interval for your angle.
You've got the right solutions so far: θ = π/12 + kπ and θ = 5π/12 + kπ. Well done!
Now, to find the interval [0, 2π), you just need to check which values of k give you angles that are within this interval. Start with k = 0 and see if the angle falls within the interval. If it does, then it's a valid solution. If not, move on to the next value of k until you reach a solution that is outside the interval.
Let's check! For k = 0:
θ = π/12 + 0π = π/12, which falls within the interval [0, 2π).
So, θ = π/12 is a valid solution.
For k = 1:
θ = π/12 + π = 13π/12, which is outside the interval [0, 2π).
So, θ = 13π/12 is NOT a valid solution.
Therefore, the only valid solution within the interval [0, 2π) is θ = π/12.
Hope that helps! Keep your angles properly intervalled next time!
To solve the equation sin(2θ) = 1/2, you can follow these steps:
Step 1: Find the reference angle
The reference angle is the angle formed between the terminal side of θ and the x-axis in standard position. Since sin is positive in both the first and second quadrants, we need to find the reference angle in either quadrant.
Using the inverse sine function, we find the reference angle:
arcsin(1/2) = π/6
Step 2: Find the values of θ in the first quadrant
Since sin is positive in the first quadrant, one solution in the first quadrant is equal to the reference angle: π/6.
Step 3: Find the values of θ in the second quadrant
To find the second quadrant solutions, subtract the reference angle from π:
π - π/6 = 5π/6
Step 4: Find all solutions in the interval [0, 2π)
To find all solutions in the given interval, you need to consider the values of θ in both the first and second quadrants and add multiples of the period to each solution.
In the first quadrant: θ = π/6
Adding the period (2π) to the solutions in the first quadrant:
θ = π/6 + 2πk, where k is any integer
In the second quadrant: θ = 5π/6
Adding the period (2π) to the solutions in the second quadrant:
θ = 5π/6 + 2πk, where k is any integer
Now, combine the solutions from both quadrants as a comma-separated list:
θ = π/6 + 2πk, 5π/6 + 2πk, where k is any integer
Therefore, the solutions to the equation sin(2θ) = 1/2 in the interval [0, 2π) are given by π/6 + 2πk, 5π/6 + 2πk, where k is any integer.