[tex]\sqrt{6}[/tex]sinx+2sin(2x-([tex]\frac{ π }{3}[/tex])=sin2x-[tex]\sqrt{3}[/tex]

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вопрос

To solve this trigonometric equation, we can start by rewriting the terms involving sine and cosine in terms of sine using the double angle formula:

sin(2x) = 2sin(x)cos(x)

So, the equation becomes:

√6sin(x) + 2(2sin(x)cos(x) - (π/3)) = 2sin(x)cos(x) - √3

Expanding the terms gives:

√6sin(x) + 4sin(x)cos(x) - 2(π/3) = 2sin(x)cos(x) - √3

Now, we can simplify the equation by grouping the terms with sin(x) and cos(x) together:

(√6 + 4cos(x))sin(x) - 2(π/3) = 2cos(x)sin(x) - √3

Next, we can use the Pythagorean identity to eliminate the cos(x) term:

cos(x) = ±√(1 - sin^2(x))

Substitute this into the equation and solve for sin(x).

(√6 + 4(±√(1 - sin^2(x)))sin(x) - 2(π/3) = 2(±√(1 - sin^2(x)))sin(x) - √3

After solving for sin(x), you can find cos(x) using cos(x) = ±√(1 - sin^2(x)). Finally, substitute sin(x) and cos(x) back into the original equation to check if it satisfies the equation.