1) (sinx+sin2x)/sin3x=1 2)4^(3+2cos2x) - 7*4^(2cos^2x)=4^0,5

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To solve the first equation:

We have (sinx + sin2x) / sin3x = 1

Expanding sin2x into sinx + cosx, we get:

(sinx + sinx + cosx) / sin3x = 1
(2sinx + cosx) / sin3x = 1
2sinx / sin3x + cosx / sin3x = 1
2cosx + cotx = 1

Now, simplify the equation further:

2cosx + cotx = 1
2cosx + (cosx/sinx) = 1
(2cosxsinx + cosx) / sinx = 1
(cos2x + cosx) / sinx = 1
cos(2x) + cos(x) = sin(x)
cos(2x) = sin(x) - cos(x)

This is the simplified form of the given equation.

To solve the second equation:

We have 4^(3 + 2cos2x) - 7 * 4^(2cos^2x) = 4^0.5

Let y = 4^(cos2x)

The equation becomes:
4^(3)y^2 - 7y = 2

Now we have a quadratic equation:
4^(cos2x)^2 - 7 * 4^(cos2x) - 2 = 0

Solve for y using the quadratic formula:

y = [7 ± √(7^2 - 4 4 (-2))] / (2 * 4^3)
y = [7 ± √(49 + 32)] / 16
y = [7 ± √81] / 16
y = [7 ± 9] / 16

Therefore, the possible solutions for y are: y = 1 or y = -1/2

Solve for 4^(cos2x) using these possible values of y to find the solutions for cos2x.