find all solutions in the interval [0,2 pi)

sin(x+(3.14/3) + sin(x- 3.14/3) =1

sin^4 x cos^2 x

Since sin (a+b) = sina cosb + cosb sina
and
sin (a-b) = sina cosb - cosb sina,
the first problem can be written
2 sin x cos (pi/3)= sin x
The solution to sin x = 1 is x = pi/2

For your other problem
sin^4 cos^2 x = sin^4 x(1 - sin^2x)=0
The solutions are sin x = 0 and sin^2 x = 1. That would correspond to x=0, pi/2, pi, and 3 pi/4

the answer is $344,000.000

Sin(x)=2π

To solve the equation sin(x + (3.14/3)) + sin(x - 3.14/3) = 1 in the interval [0, 2 pi), you can follow these steps:

1. Start by simplifying the equation using the trigonometric identities mentioned in the question.
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

2. Apply these identities to the equation:
sin(x + (3.14/3)) + sin(x - 3.14/3) = 1
sin(x)cos(3.14/3) + cos(x)sin(3.14/3) + sin(x)cos(3.14/3) - cos(x)sin(3.14/3) =1
2sin(x)cos(3.14/3) = 1

3. Simplify further:
sin(x)cos(3.14/3) = 1/2
sin(x)(1/2) = 1/2
sin(x) = 1

4. Notice that sin(x) = 1 when x = pi/2 (as mentioned in the question).

Therefore, in the interval [0, 2 pi), the solution to the equation sin(x + (3.14/3)) + sin(x - 3.14/3) = 1 is x = pi/2.

For the second part of the question, to find the solutions for sin^4(x)cos^2(x), you can follow these steps:

1. Start with the expression sin^4(x)cos^2(x).

2. Apply the identity sin^2(x) = 1 - cos^2(x):
sin^4(x)cos^2(x) = (1 - cos^2(x))^2cos^2(x)
= (1 - 2cos^2(x) + cos^4(x))cos^2(x)
= cos^2(x) - 2cos^4(x) + cos^6(x)

3. Set the expression equal to zero:
cos^2(x) - 2cos^4(x) + cos^6(x) = 0

4. Factor out cos^2(x):
cos^2(x)(1 - 2cos^2(x) + cos^4(x)) = 0

5. Solve for cos^2(x) = 0 and 1 - 2cos^2(x) + cos^4(x) = 0 separately.

For cos^2(x) = 0:
cos(x) = 0
x = pi/2

For 1 - 2cos^2(x) + cos^4(x) = 0:
Let p = cos^2(x), so the equation becomes:
p^3 - 2p^2 + p = 0

Factor out p:
p(p^2 - 2p + 1) = 0

Solve for p = 0 and p^2 - 2p + 1 = 0 separately:
For p = 0:
cos(x) = 0
x = pi/2

For p^2 - 2p + 1 = 0:
(p - 1)^2 = 0
p - 1 = 0
p = 1
cos(x) = 1
x = 0, pi, 2pi

Therefore, the solutions for sin^4(x)cos^2(x) in the interval [0, 2 pi) are x = 0, pi/2, pi, and 2pi.

As for the unrelated answer of $344,000.000, it seems to be unrelated to the given questions about trigonometric equations and does not fit in the context of the discussion.