I need to find the exact solution to:
2sin2x = sqrt(x)on the interval (0,pi)
This is how I started it, but I don't think it's correct.
sin2x = sqrt(x)/2
2sinxcosx = sqrt(x)/2
sinxcosx = sqrt(x)/4
What do I need to do now?
need to find the exact solution to:
2sin2x = sqrt(x)on the interval (0,pi)
>>
I tried this before, and spend some time on it. I could not find an exact analytical solution, the best I could do was graph it. I tricd Taylor series, infinite series for sin x, nothing worked. I am certain whomever designed the question has something in mind, some trick, but I cant decipher it.
To solve the equation 2sin2x = sqrt(x) on the interval (0, pi), it appears that you've made a good start by rewriting the equation as sin2x = sqrt(x)/2.
To continue, here's a possible approach:
1. Square both sides of the equation sin2x = sqrt(x)/2 to eliminate the square root:
(sin2x)^2 = (sqrt(x)/2)^2
Simplify the right-hand side:
sin^2(2x) = x/4
2. Recall the double-angle identity for sine: sin^2(2x) = (1 - cos(4x))/2.
3. Substitute this identity into the equation:
(1 - cos(4x))/2 = x/4
Simplify the equation:
1 - cos(4x) = x/2
4. Rearrange the equation to obtain a quadratic expression:
cos(4x) = 1 - x/2
5. Square both sides of the equation cos(4x) = 1 - x/2:
(cos(4x))^2 = (1 - x/2)^2
Simplify the right-hand side:
cos^2(4x) = 1 - x + x^2/4
6. Recall the double-angle identity for cosine: cos^2(2x) = (1 + cos(4x))/2.
7. Substitute this identity into the equation:
(1 + cos(4x))/2 = 1 - x + x^2/4
8. Multiply both sides by 2 to eliminate the fraction:
1 + cos(4x) = 2 - 2x + x^2/2
9. Rearrange the equation to obtain a quadratic expression:
cos(4x) = 1 - 2x + x^2/2
10. Solve the quadratic equation cos(4x) = 1 - 2x + x^2/2 either by factoring, completing the square, or using the quadratic formula. This will give you the values of x that satisfy the equation.
Note: It is possible that in this case, an exact analytical solution may not be possible due to the complexity of the equation. In such cases, numerical methods or approximation techniques can be used to approximate the solution.