# `sin(x) = x^2 - 2` Use Newton's method to find all roots of the equation correct to six decimal places.

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`sin(x)=x^2-2`

`f(x)=x^2-2-sin(x)=0`

`f'(x)=2x-cos(x)`

See the attached graph. From the graph, the curve of the function intersects the x-axis at `~~` -1.05 and 1.70. These can be used as initial approximations for further iteration.

`x_(n+1)=x_n-((x_n)^2-2-sin(x_n))/(2x_n-cos(x_n))`

For x_1=-1.05

`x_2=-1.05-((-1.05)^2-2-sin(-1.05))/(2(-1.05)-cos(-1.05))`

`x_2=-1.05-(1.1025-2-(-0.867423225))/(-2.1-0.497571047)`

`x_2~~-1.061578807`

carry out iteration until six digit decimal places are same

`x_3~~-1.061549775`

`x_4~~-1.061549775`

Now for x_1=1.70

`x_2=1.7-(1.7^2-2-sin(1.7))/(2*1.7-cos(1.7))`

`x_2=1.7-(2.89-2-0.99166481)/(3.4-(-0.128844494))`

`x_2~~1.72880966`

`x_3~~1.728466368`

`x_4~~1.728466319`

`x_5~~1.728466319`

Roots of the equation to six decimal places are **-1.061550 , 1.728466**