1)2cos(П/3-3x)-корень3=0 2)3+4sin(2x+1)=0

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вопрос

To solve the first equation 2cos(П/3-3x)-√3=0, we can rewrite the equation in terms of cosine and simplify it.

2cos(П/3-3x) - √3 = 0
cos(П/3-3x) = √3/2
cos(П/3-3x) = cos(П/6)

Now, we can equate the angles inside the cosine function.
П/3 - 3x = П/6
П/3 - П/6 = 3x
2П/6 - П/6 = 3x
x = П/6

Therefore, the solution to the first equation is x = П/6.

To solve the second equation 3 + 4sin(2x + 1) = 0, we can first isolate sin(2x + 1) by subtracting 3 from both sides.

4sin(2x + 1) = -3
sin(2x + 1) = -3/4

Since the sine function has a range of -1 to 1, the equation sin(2x + 1) = -3/4 has no real solutions.