Question

4sinx+корень3 sin2x=2cos2xsinx

Answer 1

$4\sin x+ \sqrt{3} \sin2x=2\cos2x\sin x\\ \\ 4\sin x+2\sqrt{3}\sin x\cos x-2\cos 2x\sin x=0\\ 2\sin x(2+\sqrt{3}\cos x-\cos2x)=0$

Произведение равно нулю, если один из множителей равен нулю

$\sin x=0\\ \\ \underline{x_1=\pi k,k \in Z}$

$2+\sqrt{3}\cos x-\cos2x=0\\ \\ 2+\sqrt{3}\cos x-2\cos^2x+1=0\\ \\ 2\cos^2x-\sqrt{3}\cos x-3=0$

Пусть  $\cos x=t(|t| \leq 1)$ тогда получаем

$2t^2-\sqrt{3}t-3=0\\ D=b^2-4ac=(-\sqrt{3})^2-4\cdot2\cdot(-3)=27\\ \\ t_1= \dfrac{-b+ \sqrt{D} }{2a} = \dfrac{\sqrt{3}+3\sqrt{3}}{2\cdot 2} =\sqrt{3}\,\, \notin [-1;1]\\ \\ \\ t_2= \dfrac{-b- \sqrt{D} }{2a} = \dfrac{\sqrt{3}-3\sqrt{3}}{2\cdot2} =- \dfrac{\sqrt{3}}{2}$

Обратная замена

$\cos x=-\dfrac{\sqrt{3}}{2} \\ \\ \underline{x_2=\pm \dfrac{5\pi}{6}+ 2 \pi n,n \in Z}$

Answer 2

To solve the equation 4sin(x) + √3sin(2x) = 2cos(2x)sin(x), we can start by simplifying the trigonometric expressions using known identities. Let's break it down step by step:

1. We know that sin(2x) = 2sin(x)cos(x) and cos(2x) = cos²(x) - sin²(x).

Substituting these identities into the equation, we have: 4sin(x) + √3 * 2sin(x) * cos(x) = 2(cos²(x) - sin²(x))sin(x)

2. Next, let's simplify the equation further: 4sin(x) + 2√3sin(x)cos(x) = 2(cos²(x) - sin²(x))sin(x)

3. Now, distribute the sin(x) term on the right side: 4sin(x) + 2√3sin(x)cos(x) = 2cos²(x)sin(x) - 2sin³(x)

4. Subtract 4sin(x) from both sides: 2√3sin(x)cos(x) = 2cos²(x)sin(x) - 2sin³(x) - 4sin(x)

5. Factor out sin(x) on the right side: 2√3sin(x)cos(x) = sin(x)(2cos²(x) - 2sin²(x) - 4)

6. Divide both sides by sin(x) (assuming sin(x) is not zero): 2√3cos(x) = 2cos²(x) - 2sin²(x) - 4

7. Apply the identity cos²(x) + sin²(x) = 1: 2√3cos(x) = 2 - 2sin²(x) - 4

8. Simplify further: 2√3cos(x) = -2sin²(x) - 2

9. Divide the entire equation by 2: √3cos(x) = -sin²(x) - 1

10. Rearrange the equation: sin²(x) + √3cos(x) + 1 = 0

Now, you have a quadratic equation in terms of sin(x) and cos(x). You can solve this equation using various methods, such as factoring, completing the square, or using the quadratic formula.