Show that the equation \( x^{2}-3=\sin x \) has a root between \( \frac{\pi}{2} \) radian And \( \pi \) radian. By taking 2 as the first approximation, evaluate this root correct to 3 decimal places.

To show that the equation \( x^{2}-3=\sin x \) has a root between \( \frac{\pi}{2} \) and \( \pi \), we can first evaluate the function \( f(x) = x^2 - 3 - \sin x \) at the endpoints of the interval. Calculating at \( x = \frac{\pi}{2} \): \[ f\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 - 3 - \sin\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} - 3 - 1 \approx 2.47 - 4 \approx -1.53. \] Now, calculating at \( x = \pi \): \[ f(\pi) = \pi^2 - 3 - \sin(\pi) = \pi^2 - 3 \approx 9.87 - 3 = 6.87. \] Since \( f\left(\frac{\pi}{2}\right) < 0 \) and \( f(\pi) > 0 \), by the Intermediate Value Theorem, there is at least one root in the interval \( \left(\frac{\pi}{2}, \pi\right) \). Now, we can take \( x_0 = 2 \) as the first approximation, and use the Newton-Raphson method to improve this approximation. The derivative \( f'(x) = 2x - \cos x \). Calculating \( f(2) \) and \( f'(2) \): \[ f(2) = 2^2 - 3 - \sin(2) \approx 4 - 3 - 0.909 = 0.091. \] \[ f'(2) = 2(2) - \cos(2) \approx 4 - (-0.416) = 4.416. \] Using the Newton-Raphson formula: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{0.091}{4.416} \approx 2 - 0.0206 \approx 1.979. \] Next, apply the method again (using \( x_1 \)): \[ f(1.979) = (1.979)^2 - 3 - \sin(1.979) \approx 3.916 - 3 - 0.914 = 0.002. \] \[ f'(1.979) = 2(1.979) - \cos(1.979) \approx 3.958 + 0.373 = 4.331. \] \[ x_2 = 1.979 - \frac{0.002}{4.331} \approx 1.979 - 0.00046 \approx 1.9785. \] Continuing this process yields: After a couple of iterations, you can find that the root converges to approximately \( 1.978 \). To check for accuracy, ensure that your final result is correct to 3 decimal places, leading us to conclude that the root of \( x^2 - 3 = \sin x \) between \( \frac{\pi}{2} \) and \( \pi \) is approximately \( x \approx 1.978 \).