Find the Roots (Zeros) 2x+cos(x)=0

Problem

2*x+cos(x)=0

Solution

1. Identify the type of equation. This is a transcendental equation because it involves both a linear term 2*x and a trigonometric term cos(x) It cannot be solved for x using basic algebraic isolation.

2. Analyze the function ƒ(x)=2*x+cos(x) Since the derivative ƒ(x)′=2−sin(x) is always positive (because sin(x) ranges from −1 to 1, the function is strictly increasing and has exactly one real root.

3. Estimate the location of the root. Since ƒ(0)=2*(0)+cos(0)=1 and ƒ*(−π/2)=2*(−π/2)+cos(−π/2)=−π≈−3.14 the Intermediate Value Theorem guarantees a root exists in the interval (−π/2,0)

4. Apply a numerical method such as Newton's Method to find the approximate value. Using the iteration formula (x_n+1)=(x_n)−ƒ((x_n))/(ƒ((x_n))′)

(x_n+1)=(x_n)−(2*(x_n)+cos((x_n)))/(2−sin((x_n)))

5. Iterate starting with an initial guess (x_0)=−0.5

(x_1)=−0.5−(2*(−0.5)+cos(−0.5))/(2−sin(−0.5))≈−0.4506

(x_2)=−0.4506−(2*(−0.4506)+cos(−0.4506))/(2−sin(−0.4506))≈−0.4501836

Final Answer

x≈−0.4501836

Want more problems? Check here!