Old School
Geometry
Level
pending
What is the sum of all the roots of following trigonometry equation:
is in radians, not degrees.
The answer is 628.32.
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1 solution
cos(X)\quad =\quad 0.29901805\ \therefore \quad X\quad =\quad 1.26713287\ \ General\quad Solution:\quad \ X\quad =\quad \pm \quad 1.2671328\quad \pm \quad 2\pi N\ where\quad N\quad =\quad 0,1,2,3,...\ \therefore \quad X\quad =\quad 1.2671328\quad +\quad 2\pi N\quad \quad \because \quad X\quad =\quad -1.2671328\quad \pm \quad 2\pi N\ We\quad will\quad Ignore\quad -1.2671328\quad -\quad 2\pi N\quad ;\quad \because \quad 1.2671328\quad -\quad 20\pi \quad \nRightarrow \quad (Do\quad Not\quad Satisfy)\quad 0\quad \le \quad X\quad \le \quad 20\pi \ \ Since\quad We\quad want\quad Sum\quad of\quad the\quad Roots\quad \ \therefore \quad Form\quad Arithmetic\quad Series\quad starting\quad with\quad N\quad =\quad 0\quad and\quad last\quad term\quad will\quad be\quad N\quad =\quad 572,\quad 573\ a=\quad \quad 1.2671328\quad ,\quad N\quad =\quad 572\quad \Rightarrow \quad (Only\quad implies)\quad +\quad 1.2671328\quad +\quad 20\pi \quad ,\quad Last\quad Term\quad l=\quad 3595.249129\ a=\quad -1.2671328,\quad N=\quad 573\quad \Rightarrow \quad 1.2671328\quad -\quad 20\pi \quad ,\quad l\quad =\quad 3598.998048\quad \ Using\quad Arithmetic\quad Formula:\ \frac { 1 }{ 2 } N(A+L)\quad +\quad \frac { 1 }{ 2 } N(A+L)\quad =\quad \frac { 1 }{ 2 } \times 572(1.2671328\quad +\quad 3595.249129)\quad +\quad \frac { 1 }{ 2 } \times 573(-1.2671328+3598.998048)\ \therefore \quad Sum\quad of\quad all\quad the\quad possible\quad roots\quad =\quad 2059353.558