Solve for sin4x=sin2x?
sin 4x = sin 2x
2sin 2x cos 2x - sin 2x = 0
sin 2x(2 cos 2x - 1) =
sin2x = 0
so 2x = 0, π, 2π, 3π, ..
x = 0, π/2, π , 3π/2, 2π for 0≤x≤2π
or cos2x = 1/2
2x = π/3, 2π-π/3, ..., the primaries
x = π/6, 5π/6
now, the period of cos2x is π, so adding or subtracting multiples of π will yield more answers
x = π/6 + π = 7π/6 --> 210°
x = 5π/6+π = 11π/6 --> 330°
then:
x = π/6, 5π/6, 7π/6, 11π/6 for 0≤x≤2π
in degrees:
x = 0,90,180,270,360 degrees
x = 30,150, 330,210 degrees
checking 150°
is sin 600 = sin300 ? yes
checking 330°
is sin 1320 = sin660? yes
Well, it seems like a mathematical problem has come my way. Let's see if we can solve it with a touch of humor!
To solve for sin(4x) = sin(2x), we can use a little trick. Remember that the sine function has a periodicity of 2π. So, we can write sin(4x) as sin(2x + 2x).
Now, let's apply a well-known trigonometric identity: sin(a + b) = sin(a)cos(b) + cos(a)sin(b). In our case, a = 2x and b = 2x. So, we have:
sin(4x) = sin(2x + 2x) = sin(2x)cos(2x) + cos(2x)sin(2x).
Since sin(2x) appears on both sides of the equation, we can cancel it out:
0 = sin(2x)cos(2x) + cos(2x)sin(2x).
Now, we're left with:
0 = 2sin(2x)cos(2x).
To satisfy this equation, we have two possibilities:
1. sin(2x) = 0, which means 2x is an integer multiple of π.
2. cos(2x) = 0, which means 2x is an odd multiple of π/2.
So, now you just need to solve for x using these conditions. Give it a try, and remember, math can be fun, just like a clown juggling numbers!
To solve the equation sin(4x) = sin(2x), we can use the trigonometric identity, which states that sin(a) = sin(b) if either a = b or a + b = 180 degrees (or any multiple of 180 degrees).
So we have two possible cases:
Case 1: 4x = 2x
Simplifying, we get:
2x = 0
Dividing both sides by 2, we get:
x = 0
Case 2: 4x + 2x = 180 degrees
Simplifying, we get:
6x = 180 degrees
Dividing both sides by 6, we get:
x = 30 degrees
Therefore, the solutions for sin(4x) = sin(2x) are x = 0 and x = 30 degrees.
To solve the equation sin(4x) = sin(2x), we can use the identity that sin(2θ) = 2sin(θ)cos(θ).
First, let's apply this identity to the equation:
sin(4x) = sin(2x)
2sin(2x)cos(2x) = sin(2x)
Now, we have sin(2x) on both sides of the equation.
There are two cases to consider:
Case 1: sin(2x) = 0
If sin(2x) = 0, this means that either sin(2x) = 0 or cos(2x) = 0.
If sin(2x) = 0, this implies that 2x = kπ, where k is an integer. Therefore, x = kπ/2.
If cos(2x) = 0, this implies that 2x = (2k+1)π/2, where k is an integer. Therefore, x = (2k+1)π/4.
Case 2: sin(2x) ≠ 0
In this case, we can divide both sides of the equation by sin(2x):
2sin(2x)cos(2x) / sin(2x) = sin(2x) / sin(2x)
This simplifies to:
2cos(2x) = 1
Dividing both sides by 2:
cos(2x) = 1/2
Now, we can solve for 2x:
2x = π/3 + 2kπ or 2x = 5π/3 + 2kπ
Divide both sides by 2:
x = π/6 + kπ or x = 5π/6 + kπ
In summary, the solutions to the equation sin(4x) = sin(2x) are:
- For sin(2x) = 0: x = kπ/2 or x = (2k+1)π/4
- For sin(2x) ≠ 0: x = π/6 + kπ or x = 5π/6 + kπ, where k is an integer.