Find x where 0<=x<=2pi sin x - cot^2 x = 1 Anyone know how to do these things??

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Find x where 0<=x<=2pi sin x - cot^2 x = 1 Anyone know how to do these things??

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Answer 1 · 2018-05-21T22:12:34

We can rewrite:

$sin x = 1 + {cot}^{2} x$ $sinx = 1 + cot^2x$

Now recall that ${cot}^{2} x + 1 = {csc}^{2} x$ $cot^2x + 1 = csc^2x$ .

$sin x = {csc}^{2} x$ $sinx = csc^2x$

$sin x - \frac{1}{{sin}^{2} x} = 0$ $sinx -1/sin^2x = 0$

$\frac{{sin}^{3} x - 1}{{sin}^{2} x} = 0$ $(sin^3x - 1)/sin^2x= 0$

${sin}^{3} x = 1$ $sin^3x =1$

$sin x = 1$ $sinx = 1$

$x = \frac{π}{2}$ $x = pi/2$

We can confirm graphically our answer is correct.

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Hopefully this helps!

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Find x where 0<=x<=2pi sin x - cot^2 x = 1 Anyone know how to do these things??

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