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Verify, by calculation that the roots of the equation $\csc x = \frac{1}{2}x + 1$, where x is n radians, has roots in the interval $0

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    So, what's your doubt? – 2017-01-24
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    I tried solving this a lot but i got very confused about which root to take and i've tried and tried and tried but i just couldn't solve it... please help me! – 2017-01-24

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Please see here: enter image description here

The root $x=0.797$ lies in $(0,\frac {\pi}{2}) $ verifying the result. Hope it helps.

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The root may be found with Newton's method:

$$0=x+2-2\csc(x)$$

$$x_{n+1}=x_n-\frac{x_n+2-2\csc(x_n)}{1+2\csc(x_n)\cot(x_n)}$$

With $x_0=0.8$, we get

$x_1=0.7967677637$

$x_2=0.7967790659$

$x_3=0.7967790660$

Which is the solution out ten places.

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    ??? Why the downvote? Is it not a tad bit unreasonable? – 2017-01-24