sin^2(2x)=2sinxcosx. Find all solutions to each equation in the interval [0, 2pi)
So I started off changing 2sinxcosx = sin(2x), and my equation ended as
sin^2(2x) = sin(2x).
I subtracted sin(2x) by both sides and factored out sin(2x).
my equation ended like so:
sin^2(2x) - sin(2x) = sin(2x)*(sin(2x)-1)=0.
I was told to substitute 2x with some other variable to make it easier for me to solve this. So,
sin(2x)*(sin(2x)-1) = sin(y)*(sin(y)-1),
and siny = 0,1.
At this point, I am stuck because I don't know what to do when I change y back to 2x.
I would really appreciate your help with this!
p.s. Rest in peace, Mrs. Sue. She has saved my grades countless times and other students using this website as well.
so, you have sin(2x) = 0 or sin(2x) = 1
You know that sin(0) = 0 and sin(π/2) = 1
So, 2x = 0 or 2x = π/2
Thus, x = 0 or x = π/4
The period of sin 2x is π, so the complete solution is
x = 0 + kπ
x = π/4 + kπ
for any integer k.
Thank you so much for your help!
To solve the equation sin^2(2x) = 2sin(x)cos(x), let's start from the point where you factor out sin(2x):
sin^2(2x) - sin(2x) = sin(2x)*(sin(2x) - 1) = 0.
Now, substitute y = 2x into the equation:
sin(y)*(sin(y) - 1) = 0.
To find the solutions, we need to solve sin(y) = 0 and sin(y) - 1 = 0 separately.
1. for sin(y) = 0:
In the interval [0, 2π), sin(y) = 0 has solutions at y = 0, π, and 2π.
2. for sin(y) - 1 = 0:
Rearrange the equation: sin(y) = 1.
In the interval [0, 2π), sin(y) = 1 has a solution at y = π/2.
Now, we substitute y back into 2x:
1. y = 0 gives 2x = 0, which means x = 0.
2. y = π gives 2x = π, which means x = π/2.
3. y = 2π gives 2x = 2π, which means x = π.
Therefore, the solutions to the original equation in the interval [0, 2π) are x = 0, x = π/2, and x = π.
I'm sorry to hear about Mrs. Sue. She must have been a wonderful teacher.
To solve the equation sin^2(2x) - sin(2x) = 0, you correctly factored out sin(2x) to get sin(2x) * (sin(2x) - 1) = 0.
Now, you need to solve the two equations sin(2x) = 0 and sin(2x) - 1 = 0 separately.
For sin(2x) = 0, you can set 2x = arcsin(0), which gives you 2x = 0 or π. Thus, x = 0 or x = π/2.
For sin(2x) - 1 = 0, you can add 1 to both sides to get sin(2x) = 1. Now, we have two possibilities:
1) Determine the angle whose sine is 1: 2x = arcsin(1), which gives you 2x = π/2. Therefore, x = π/4.
2) Determine the reference angle: If sin(θ) = 1, the reference angle is π/2. However, in the interval [0, 2π), we need to find all values of 2x that are equivalent to π/2. These values are π/4 and 5π/4. Therefore, x = π/8 and x = 5π/8.
Putting all the solutions together, we have: x = 0, π/2, π/4, π/8, and 5π/8.
I hope this helps! If you have any further questions, please feel free to ask.