1 + sin7x= (cos3x/2 + sin3x/2)^2

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вопрос

To solve this equation, we can simplify the right side first:

(cos3x/2 + sin3x/2)^2 = (cos^2(3x/2) + 2cos(3x/2)sin(3x/2) + sin^2(3x/2)
= (1 + 2 cos(3x/2)sin(3x/2)
= 1 + sin(3x)

Now our equation becomes:

1 + sin(7x) = 1 + sin(3x)

Subtracting 1 from both sides, we get:

sin(7x) = sin(3x)

Now we need to find the solutions for this trigonometric equation.

sin(7x) = sin(3x)

7x = 3x + 2n(pi) or 7x = (pi) - 3x + 2n(pi
4x = 2n(pi) or 10x = (pi) + 2n(pi)

x = n(pi)/2 or x = (pi/10) + n(pi/5)

So, the solutions to the equation is x = n(pi)/2 or x = (pi/10) + n(pi/5) for any integer n.