Question

2 sin^2x+2sinx = sqrt(3)+ sqrt(3)sinx

Answer 1

$2\sin^2x+2\sin x= \sqrt{3} +\sqrt{3}\sin x \\ 2\sin^2 x+2\sin x-(\sqrt{3}+\sqrt{3}\sin x)=0 \\ 2\sin x(\sin x+1)-\sqrt{3}(1+\sin x)=0 \\ (\sin x+1)(2\sin x-\sqrt{3})=0 \\ \left[\begin{array}{ccc}\sin x=-1\\\sin x= \frac{\sqrt{3}}{2} \end{array}\right\to \left[\begin{array}{ccc}x_1=- \frac{\pi}{2} +2 \pi k,k \in Z\\ x_2=(-1)^k\cdot \frac{\pi}{3}+ \pi k,k \in Z \end{array}\right$

Answer 2

To solve the equation $2\sin^2(x) + 2\sin(x) = \sqrt{3} + \sqrt{3}\sin(x)$, we can follow these steps:

1. Move all terms to one side of the equation to set it to zero: $2\sin^2(x) + 2\sin(x) - \sqrt{3} - \sqrt{3}\sin(x) = 0$

2. Combine like terms: $2\sin^2(x) + (2 - \sqrt{3})\sin(x) - \sqrt{3} = 0$

3. Let's make a substitution to simplify this equation. We can set $y = \sin(x)$, which transforms the equation into a quadratic equation in terms of $y$: $2y^2 + (2 - \sqrt{3})y - \sqrt{3} = 0$

4. Solve this quadratic equation for $y$: You can use the quadratic formula to solve for (y: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 2$, $b = 2 - \sqrt{3}$, and $c = -\sqrt{3}$.

Plug these values into the quadratic formula:

$y = \frac{-(2 - \sqrt{3}) \pm \sqrt{(2 - \sqrt{3})^2 - 4(2)(-\sqrt{3})}}{2(2)}$

Now, calculate the values for $y$:

$y_1 = \frac{-(2 - \sqrt{3}) + \sqrt{(2 - \sqrt{3})^2 - 4(2)(-\sqrt{3})}}{4}$

$y_2 = \frac{-(2 - \sqrt{3}) - \sqrt{(2 - \sqrt{3})^2 - 4(2)(-\sqrt{3})}}{4}$

Now that you have the values for $y$, you can find the corresponding values for $x$ by taking the arcsine of $y$:

$x_1 = \arcsin(y_1)$ $x_2 = \arcsin(y_2)$

Please note that you may end up with two or more solutions for $x$ depending on the values of $y$. Calculate

Answer 3

To solve the equation 2sin^2(x) + 2sin(x) = √3 + √3sin(x), you can start by moving all terms to one side of the equation to set it to zero:

2sin^2(x) + 2sin(x) - √3 - √3sin(x) = 0

Now, let's rearrange the terms and combine like terms:

2sin^2(x) - √3sin(x) + 2sin(x) - √3 = 0

Next, factor out common terms if possible. In this case, you can factor out sin(x):

sin(x)(2sin(x) - √3) + (√3 - 2) = 0

Now you have a quadratic equation in terms of sin(x):

2sin(x)^2 - √3sin(x) + (√3 - 2) = 0

To solve this quadratic equation for sin(x), you can use the quadratic formula:

sin(x) = [-b ± √(b^2 - 4ac)] / (2a)

In this case, a = 2, b = -√3, and c = √3 - 2. Plugging these values into the quadratic formula:

sin(x) = [√3 ± √((-√3)^2 - 4 * 2 * (√3 - 2))] / (2 * 2)

Now, simplify the expression under the square root:

sin(x) = [√3 ± √(3 + 16(2 - √3))] / 4

sin(x) = [√3 ± √(3 + 32 - 48√3)] / 4

sin(x) = [√3 ± √(35 - 48√3)] / 4

Now, you have two possible solutions for sin(x):

1. sin(x) = [√3 + √(35 - 48√3)] / 4
2. sin(x) = [√3 - √(35 - 48√3)] / 4

You can further simplify these solutions if needed, but these are the exact solutions for sin(x).