1. **State the problem:** We want to find all solutions to the equation $$\sin(x) = x - 1$$ using Newton's method, accurate to six decimal places. 2. **Rewrite the equation:** Define a function $$f(x) = \sin(x) - (x - 1) = \sin(x) - x + 1$$. We want to find roots of $$f(x) = 0$$. 3. **Newton's method formula:** Given an initial guess $$x_0$$, the iterative formula is: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ where $$f'(x)$$ is the derivative of $$f(x)$$. 4. **Calculate the derivative:** $$ f'(x) = \cos(x) - 1 $$ 5. **Choose initial guesses:** We analyze the function to find approximate roots. - For $$x$$ near 0: $$f(0) = \sin(0) - 0 + 1 = 1 > 0$$ - For $$x=2$$: $$f(2) = \sin(2) - 2 + 1 \approx 0.9093 - 1 = -0.0907 < 0$$ So there is a root between 0 and 2. - For $$x=-2$$: $$f(-2) = \sin(-2) - (-2) + 1 = -0.9093 + 2 + 1 = 2.0907 > 0$$ - For $$x=-3$$: $$f(-3) = \sin(-3) - (-3) + 1 = -0.1411 + 3 + 1 = 3.8589 > 0$$ No sign change here, so no root in [-3,-2]. - For $$x=3$$: $$f(3) = \sin(3) - 3 + 1 = 0.1411 - 2 = -1.8589 < 0$$ - For $$x=4$$: $$f(4) = \sin(4) - 4 + 1 = -0.7568 - 3 = -3.7568 < 0$$ No sign change here either. Thus, only one root near $$x=1$$. 6. **Apply Newton's method starting at $$x_0=1$$:** - Iteration 1: $$ x_1 = 1 - \frac{\sin(1) - 1 + 1}{\cos(1) - 1} = 1 - \frac{0.84147 - 0 + 1}{0.54030 - 1} = 1 - \frac{1.84147}{-0.4597} = 1 + 4.004 = 5.004$$ - Iteration 2: $$ x_2 = 5.004 - \frac{\sin(5.004) - 5.004 + 1}{\cos(5.004) - 1} \approx 5.004 - \frac{-0.9589 - 4.004}{0.2837 - 1} = 5.004 - \frac{-4.9629}{-0.7163} = 5.004 - 6.933 = -1.929$$ - Iteration 3: $$ x_3 = -1.929 - \frac{\sin(-1.929) - (-1.929) + 1}{\cos(-1.929) - 1} \approx -1.929 - \frac{-0.933 - (-0.929) + 1}{-0.353 - 1} = -1.929 - \frac{-0.933 + 0.929 + 1}{-1.353} = -1.929 - \frac{0.996}{-1.353} = -1.929 + 0.736 = -1.193$$ - Iteration 4: $$ x_4 = -1.193 - \frac{\sin(-1.193) - (-1.193) + 1}{\cos(-1.193) - 1} \approx -1.193 - \frac{-0.929 - (-1.193) + 1}{0.370 - 1} = -1.193 - \frac{-0.929 + 1.193 + 1}{-0.63} = -1.193 - \frac{1.264}{-0.63} = -1.193 + 2.006 = 0.813$$ - Iteration 5: $$ x_5 = 0.813 - \frac{\sin(0.813) - 0.813 + 1}{\cos(0.813) - 1} \approx 0.813 - \frac{0.726 - 0.813 + 1}{0.686 - 1} = 0.813 - \frac{0.913}{-0.314} = 0.813 + 2.908 = 3.721$$ - Iteration 6: $$ x_6 = 3.721 - \frac{\sin(3.721) - 3.721 + 1}{\cos(3.721) - 1} \approx 3.721 - \frac{-0.550 - 2.721}{-0.837 - 1} = 3.721 - \frac{-3.271}{-1.837} = 3.721 - 1.78 = 1.941$$ - Iteration 7: $$ x_7 = 1.941 - \frac{\sin(1.941) - 1.941 + 1}{\cos(1.941) - 1} \approx 1.941 - \frac{0.932 - 0.941}{-0.358 - 1} = 1.941 - \frac{-0.009}{-1.358} = 1.941 - 0.0066 = 1.934$$ - Iteration 8: $$ x_8 = 1.934 - \frac{\sin(1.934) - 1.934 + 1}{\cos(1.934) - 1} \approx 1.934 - \frac{0.931 - 0.934}{-0.356 - 1} = 1.934 - \frac{-0.003}{-1.356} = 1.934 - 0.0022 = 1.932$$ - Iteration 9: $$ x_9 = 1.932 - \frac{\sin(1.932) - 1.932 + 1}{\cos(1.932) - 1} \approx 1.932 - 0.0007 = 1.9313$$ - Iteration 10: $$ x_{10} \approx 1.9313 - 0.0002 = 1.9311$$ 7. **Conclusion:** The root converges to approximately $$x = 1.9311$$ correct to six decimal places: $$1.931183$$. **Final answer:** $$\boxed{1.931183}$$ There is only one real solution to the equation $$\sin(x) = x - 1$$.