Question

Arcsin(x^2-x)=arcsin(4x+6)

Answer 1

Arcsin(x^2-x)=arcsin(4x+6)/ *sina
sin(arcsin(x^2-x))=sin(arcsin(4x+6))
x^2-x=4x+6
x^2-5x-6=0
d=5^2-4*1*(-6)=49
x1,2=(5+-7)/2=-1;6

Answer 2

To solve the equation arcsin(x^2 - x) = arcsin(4x + 6), we need to isolate x. However, it's important to note that the arcsine function has a limited domain and range, which restricts the values of x we can consider.

First, let's analyze the domain and range of the arcsin function. The domain of arcsin is -1 ≤ x ≤ 1, and the range is -π/2 ≤ y ≤ π/2.

Now, let's solve the equation step by step:

Step 1: Set the angles equal to each other. arcsin(x^2 - x) = arcsin(4x + 6)

Step 2: Remove the arcsin function by taking the sine of both sides. sin(arcsin(x^2 - x)) = sin(arcsin(4x + 6))

Step 3: Apply the identity sin(arcsin(y)) = y. x^2 - x = 4x + 6

Step 4: Rearrange the equation and set it equal to zero. x^2 - x - 4x - 6 = 0 x^2 - 5x - 6 = 0

Step 5: Factorize the quadratic equation or use the quadratic formula to solve for x. (x - 6)(x + 1) = 0

Setting each factor equal to zero gives us: x - 6 = 0 or x + 1 = 0

Solving these equations gives us: x = 6 or x = -1

Step 6: Check the solutions in the original equation to ensure they are valid within the domain and range of arcsin.

For x = 6: arcsin((6)^2 - 6) = arcsin(4(6) + 6) arcsin(36 - 6) = arcsin(30)

Since 30 is outside the range of arcsin, x = 6 is not a valid solution.

For x = -1: arcsin((-1)^2 - (-1)) = arcsin(4(-1) + 6) arcsin(1 + 1) = arcsin(2)

Since 2 is within the range of arcsin, x = -1 is a valid solution.

Therefore, the only valid solution to the equation arcsin(x^2 - x) = arcsin(4x + 6) is x = -1.