Find The Solution Set For X Given 2x^2 X -10 =0

What is the general solution of sin^2(x) - cos(2x) = 2 - sin(2x) given that 3cosx is not equal to 2sin x?

[math]\sin(x)^2 -\cos(2x) = 2 - \sin(2x)[/math]thus:[math]\sin(x)^2 -\cos(x)^2 + \sin(x)^2 = 2 - 2\sin(x)\cos(x)[/math]or:[math]2\sin(x)^2 -\cos(x)^2 = 2 - 2\sin(x)\cos(x)[/math]also[math]-2\sin(x)^2 -2\cos(x)^2 = -2[/math]adding the last two equations:[math]-3\cos(x)^2 = - 2\sin(x)\cos(x)[/math]or:[math]\cos(x) (2\sin(x) - 3\cos(x)) = 0[/math]So if [math]3\cos(x) \neq 2 \sin(x)[/math], we must have:[math]\cos(x) = 0[/math], and consequently [math]\sin(x) = 1[/math] or [math]\sin(x) = -1[/math]which both cases satisfy the original equation:[math]2\sin(x)^2 -\cos(x)^2 = 2 - 2\sin(x)\cos(x)[/math][math]2*1^2 -0^2 = 2 - 2*1*0[/math] and [math]2*(-1)^2 -0^2 = 2 - 2*(-1)*0[/math]Therefore the only solution is [math]\cos(x) = 0[/math] or:[math]x = (n \pi + \frac{\pi}{2}), \forall n \in \mathbb{Z}.[/math]

Given that [math](x+iy)^2=12i-5, x, y[/math] real, find the set of all  possible [math]x+iy[/math].  Then solve the equation [math]z^2+4z=12i-9[/math].

(x + iy)^2 = 12i -5 x^2 - y^2  + 2ixy = 12i - 5 comparing real and imaginary values x^2 - y^2 = -5        (real)     (equation - 1)and 2xy = 12          (imaginary)solving the imaginary part xy = 6  (equation - 2 )puting  y = 6/x in equation 1 we get x^4 + 5x^2 -36 = 0 (x^2 + 9)*(x^2-4) = 0 that means x = +-3i or x = +-2 since both numbers are real x = +-2 hence putting this in equation - 2we get y = +-3 hence (x,y) = (2,3) or (-2,-3)Now looking at the second half of the question....z^2 + 4z = 12i - 9 adding subtracting +4 on both sidesz^2 + 4z +4 = 12i - 5 from first part of the question (x + iy)^2 = 12i -5 hence we could write z^2 + 4z + 4 = (x + iy)^2z^2 + 4z +4 is also equal to (z+2)^2 hence (z+2)^2 = (x+iy)^2now this means z + 2 = x + iy z = (x-2) + iy since (x,y) = (2,3) or (-2,-3)z = ((2-2) +3i) or ((-2-2) - 3i)z = 3i or (-4-3i)

What is the solution to 2x/(2x^2+5x+2) < 1/(x+1)?

[math]\dfrac{2x}{2x^2+5x+2}<\dfrac{1}{x+1}[/math]Factoring:[math]\dfrac{2x}{(2x+1)(x+2)}<\dfrac{1}{x+1}[/math]Since denominators cannot be 0, we know that the inequality is undefined for the x values [math]-2,-1,-\frac{1}{2}[/math]If this was an equation, we would multiply by all the factors in the denominators, which means we would multiply by:[math](2x+1)(x+2)(x+1)[/math]We can do the same thing here, but we have to know if we are multiplying by something negative or something positive. If [math]x<-2[/math], we would be multiplying by something negative, because all 3 items multiplied would be negative. Similarly, if [math]-2 -\frac{1}{2}[/math], we would be multiplying by something positive. Therefore, I will break this up into cases. When we are multiplying by something negative, we flip the inequality. When we are multiplying by something positive, the inequality remains the same:[math]\begin{cases}2x(x+1)>(2x+1)(x+2) &\text{for }x \in (-\infty,-2)\cup(-1,-\frac{1}{2}) \\ 2x(x+1)<(2x+1)(x+2) &\text{for }x \in (-2,-1)\cup(-\frac{1}{2},\infty) \end{cases}[/math]I will be careful not to perform any multiplication or division by a negative, so I will treat those inequalities as equations for solving purposes.[math]2x(x+1)=(2x+1)(x+2)[/math][math]2x^2+2x=2x^2+5x+2[/math][math]0=3x+2[/math][math]-2=3x[/math][math]-\frac{2}{3}=x[/math]Therefore, the solved version of the inequalities ends up as follows:[math]\begin{cases} -\frac{2}{3}>x &\text{for }x \in (-\infty,-2)\cup(-1,-\frac{1}{2}) \\ -\frac{2}{3}x[/math] and [math]x<-2[/math][math]-\frac{2}{3}>x[/math] and [math]-1-\frac{1}{2}[/math]The first of the 4 possibilities gives us [math]x<-2.[/math]The second of the 4 possibilities gives us [math]-1-\frac{1}{2}[/math]Therefore, the solution set is the union of those 4:[math]x \in (-\infty,-2)\cup(-1,-\frac{2}{3})\cup(-\frac{1}{2},\infty)[/math]

What is the solution set of the following equation: X 2 = x -?

It is very difficult to make any headway, since it is very unclear what the question is supposed to be.Two reasonable interpretations of the LHS of the equation would be X(2) and X^2. However, conventionally we would write 2X, so I am going to be assumed the LHS should be X^2. The RHS has x - (assuming the question mark is being used as a punctuation mark and not a mathematical symbol). This has no reasonable interpretation, but seeing as the use of (-) as a binary operator is really far-fetched here, I am going to assume it means (-) as as a unary operator, hence -x.There are an infinite number of solutions, if we assume no relation between x and X, but I am going to assume for the sake of making some headway that x = X.Then, we have x^2 = -x, which has two solutions in the complex numbers (Fundamental Theorem of Algebra), and if we re-arrange and factor, we see these values are very easy to find:x^2 = -xx^2 + x = 0x(x+1) = 0Hence, the solutions are x=0 and x=-1, and we see both work as 0^2 = -0, and (-1)^2 = -(-1). However, let us revisit the assumptions for how (-) was used. This might suggest that the X (2) should be interpreted as 2X, hence let us address that case, again considering x = X.Then, we have 2x = -x, which has 1 solution in the complex numbers by the Fundamental Theorem of Algebra, which is easy enough to find by re-arranging, namely 3x = 0, hence x = 0.Seeing as this question is markedly unclear, and in the two “reasonable” interpretations, x = X = 0 is the only element in the intersection of the solution sets, I would say the solution set is {x = 0 | x = X}, where “|” is to be understood in plain English as “given” (as one would see in conditional probability).