Sinx(2sinx-1)+√3sinx+sin4π/3=0

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вопрос

To solve the given equation:
Sinx(2sinx-1)+√3sinx+sin(4π/3)=0

Let's first simplify the equation by expanding sin(4π/3) and then combining like terms:

sinx(2sinx-1) + √3sinx + sin(4π/3)
= 2sin^2(x) - sinx + √3sinx -√3/2
= 2sin^2(x) + (√3-1)sinx -√3/2

Now, we have a quadratic equation in terms of sinx:

2sin^2(x) + (√3-1)sinx -√3/2 = 0

Let's substitute sinx = a in this quadratic equation:

2a^2 + (√3-1)a -√3/2 = 0

Solving this quadratic equation, we can find the values of a which will give us the solutions for sinx.