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The real root of the equation 5x − 2cosx −1 = 0 (up to two decimal accuracy) is _______
(Important - Enter only the numerical value in the answer)
Correct answer is between '0.53,0.56'. Can you explain this answer?
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The real root of the equation 5x − 2cosx −1 = 0 (up to two...
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The real root of the equation 5x − 2cosx −1 = 0 (up to two...
Understanding the Equation
The equation given is:
5x - 2cos(x) - 1 = 0.
This is a transcendental equation, meaning it involves both algebraic and trigonometric functions, making it non-linear and complex to solve analytically.
Identifying the Root
To find the real root of the equation, we can use numerical methods such as the bisection method or Newton-Raphson method. The correct answer is known to be between 0.53 and 0.56.
Evaluating the Function
1. Function Definition: Define the function as:
- f(x) = 5x - 2cos(x) - 1.
2. Evaluate at Endpoints: Check the values of f(x) at the endpoints:
- f(0.53) = 5(0.53) - 2cos(0.53) - 1.
- f(0.56) = 5(0.56) - 2cos(0.56) - 1.
3. Sign Changes: Look for a sign change between f(0.53) and f(0.56). If there is a sign change, it indicates a root exists in that interval.
Using Numerical Methods
- By applying the bisection method, you can find the root iteratively:
- Start with the midpoint of the interval (0.53 + 0.56)/2.
- Evaluate f(midpoint) and adjust the interval based on the sign.
Final Result
After performing these evaluations and iterations, the real root of the equation, rounded to two decimal places, is approximately 0.54.
The equation given is:
5x - 2cos(x) - 1 = 0.
This is a transcendental equation, meaning it involves both algebraic and trigonometric functions, making it non-linear and complex to solve analytically.
Identifying the Root
To find the real root of the equation, we can use numerical methods such as the bisection method or Newton-Raphson method. The correct answer is known to be between 0.53 and 0.56.
Evaluating the Function
1. Function Definition: Define the function as:
- f(x) = 5x - 2cos(x) - 1.
2. Evaluate at Endpoints: Check the values of f(x) at the endpoints:
- f(0.53) = 5(0.53) - 2cos(0.53) - 1.
- f(0.56) = 5(0.56) - 2cos(0.56) - 1.
3. Sign Changes: Look for a sign change between f(0.53) and f(0.56). If there is a sign change, it indicates a root exists in that interval.
Using Numerical Methods
- By applying the bisection method, you can find the root iteratively:
- Start with the midpoint of the interval (0.53 + 0.56)/2.
- Evaluate f(midpoint) and adjust the interval based on the sign.
Final Result
After performing these evaluations and iterations, the real root of the equation, rounded to two decimal places, is approximately 0.54.
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The real root of the equation 5x − 2cosx −1 = 0 (up to two decimal accuracy) is _______(Important - Enter only the numerical value in the answer)Correct answer is between '0.53,0.56'. Can you explain this answer?
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