Asked by kwack
4.2
18. show that the equation 2x-1-sinx=0 has exactly one real root.
18. show that the equation 2x-1-sinx=0 has exactly one real root.
Answers
Answered by
Damon
sin x = 2x-1
2x = sin x - 1
range of sin x is -1 to + 1
so 2x is between -2 and 0
so x is between -1 and 0
2x = sin x - 1
range of sin x is -1 to + 1
so 2x is between -2 and 0
so x is between -1 and 0
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