1)sinx = 1 - 2sin²x 2)√2 cos²7x - cos7x = 0

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1) Let's simplify the given equation sinx = 1 - 2sin²x:

Rearranging the terms, we get:
2sin²x + sinx - 1 = 0

This is a quadratic equation in sinx. We can solve this by factorizing or using the quadratic formula.

By factoring,
(2sinx - 1)(sinx + 1) = 0

This gives sinx = 1/2 or sinx = -1.

Therefore, the solutions are x = π/6, 5π/6, and 3π/2.

2) Let's simplify the given equation √2 cos²(7x) - cos(7x) = 0:

Let's make a substitution to make the equation easier to solve:
Let y = cos(7x)

The given equation becomes:
√2y² - y = 0
y(√2y - 1) = 0

So, the solutions for y are y = 0 and y = 1/√2.

Now, we can find the solutions for x by solving for y in terms of cos(7x):
cos(7x) = 0 and cos(7x) = 1/√2

For cos(7x) = 0:
7x = π/2 + nπ
x = π/14 + nπ/7

For cos(7x) = 1/√2:
7x = π/4 + 2nπ or 7x = 7π/4 + 2nπ
x = π/28 + 2nπ/7 or x = π/4 + 2nπ/7

Therefore, the solutions are x = π/14, 2π/14, 3π/14, 4π/14, 5π/14, 6π/14, π/4, π/4 + π/7, π/4 + 2π/7, 7π/14, 15π/14, and 29π/14.