Studdy AI Logo
Math  /  Algebra

Question10. xy=(x)sin⁡−12x\frac{x}{y}=(x)^{\sin ^{-1} 2 x}

Studdy Solution

STEP 1

1. The equation involves trigonometric and exponential functions.
2. We assume sin⁡−1 \sin^{-1} denotes the inverse sine function, also known as arcsine.
3. The goal is to solve for x x in terms of y y , or to find conditions on x x and y y .

STEP 2

1. Analyze the domain of the inverse sine function.
2. Simplify the equation if possible.
3. Solve for x x in terms of y y .

STEP 3

First, consider the domain of the inverse sine function, sin⁡−1(2x) \sin^{-1}(2x) . The function sin⁡−1(u) \sin^{-1}(u) is defined for −1≤u≤1 -1 \leq u \leq 1 . Therefore, we have:
−1≤2x≤1 -1 \leq 2x \leq 1
Divide the entire inequality by 2 to solve for x x :
−12≤x≤12 -\frac{1}{2} \leq x \leq \frac{1}{2}

STEP 4

Given the equation xy=xsin⁡−1(2x) \frac{x}{y} = x^{\sin^{-1}(2x)} , we want to simplify or manipulate it.
First, let's consider the case when x=0 x = 0 :
If x=0 x = 0 , then the left-hand side becomes 0y=0 \frac{0}{y} = 0 , and the right-hand side becomes 0sin⁡−1(0)=00 0^{\sin^{-1}(0)} = 0^0 , which is indeterminate. Thus, x≠0 x \neq 0 .
Assuming x≠0 x \neq 0 , we can take the natural logarithm of both sides to simplify:
ln⁡(xy)=ln⁡(xsin⁡−1(2x)) \ln\left(\frac{x}{y}\right) = \ln\left(x^{\sin^{-1}(2x)}\right)
Using the logarithmic identity ln⁡(ab)=bln⁡(a) \ln(a^b) = b\ln(a) , we have:
ln⁡(x)−ln⁡(y)=sin⁡−1(2x)⋅ln⁡(x) \ln(x) - \ln(y) = \sin^{-1}(2x) \cdot \ln(x)

STEP 5

Rearrange the equation to isolate terms involving x x :
ln⁡(x)−sin⁡−1(2x)⋅ln⁡(x)=ln⁡(y) \ln(x) - \sin^{-1}(2x) \cdot \ln(x) = \ln(y)
Factor out ln⁡(x) \ln(x) from the left-hand side:
ln⁡(x)⋅(1−sin⁡−1(2x))=ln⁡(y) \ln(x) \cdot (1 - \sin^{-1}(2x)) = \ln(y)

STEP 6

Solve for x x in terms of y y :
Assuming 1−sin⁡−1(2x)≠0 1 - \sin^{-1}(2x) \neq 0 , divide both sides by 1−sin⁡−1(2x) 1 - \sin^{-1}(2x) :
ln⁡(x)=ln⁡(y)1−sin⁡−1(2x) \ln(x) = \frac{\ln(y)}{1 - \sin^{-1}(2x)}
Exponentiate both sides to solve for x x :
x=eln⁡(y)1−sin⁡−1(2x) x = e^{\frac{\ln(y)}{1 - \sin^{-1}(2x)}}
This expression gives x x in terms of y y and sin⁡−1(2x) \sin^{-1}(2x) . Further simplification may not be possible without additional constraints or numerical methods.
The solution for x x in terms of y y is:
x=eln⁡(y)1−sin⁡−1(2x) x = e^{\frac{\ln(y)}{1 - \sin^{-1}(2x)}}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord