How do you solve ${csc}^{2} x - csc x + 9 = 11$ $csc^2x-cscx+9=11$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2016-09-19T14:55:50

$x = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$ $x = (pi) / (6), (5 pi) / (6), (3 pi) / (2)$

We have: ${csc}^{2} (x) - csc (x) + 9 = 11$ $csc^(2)(x) - csc(x) + 9 = 11$ ; $0 \leq x \leq 2 π$ $0 leq x leq 2 pi$

$\Rightarrow {csc}^{2} (x) - csc (x) - 2 = 0$ $=> csc^(2)(x) - csc(x) - 2 = 0$

$\Rightarrow {csc}^{2} (x) + csc (x) - 2 csc (x) - 2 = 0$ $=> csc^(2)(x) + csc(x) - 2 csc(x) - 2 = 0$

$\Rightarrow csc (x) (csc (x) + 1) - 2 (csc (x) + 1) = 0$ $=> csc(x) (csc(x) + 1) - 2 (csc(x) + 1) = 0$

$\Rightarrow (csc (x) + 1) (csc (x) - 2) = 0$ $=> (csc(x) + 1 ) (csc(x) - 2) = 0$

$\Rightarrow csc (x) + 1 = 0$ $=> csc(x) + 1 = 0$

$\Rightarrow csc (x) = - 1$ $=> csc(x) = - 1$

$\Rightarrow x = {csc}^{- 1} (- 1)$ $=> x = csc^(- 1)(- 1)$

$\Rightarrow x = \frac{3 π}{2}$ $=> x = (3 pi) / (2)$

or

$\Rightarrow csc (x) - 2 = 0$ $=> csc(x) - 2 = 0$

$\Rightarrow csc (x) = 2$ $=> csc(x) = 2$

$\Rightarrow x = {csc}^{- 1} (2)$ $=> x = csc^(- 1)(2)$

$\Rightarrow x = \frac{π}{6}$ $=> x = (pi) / (6)$

Then, the domain given is $0 \leq x \leq 2 π$ $0 leq x leq 2 pi$ :

$\Rightarrow x = \frac{π}{6}, \frac{3 π}{2}, π - \frac{π}{6}$ $=> x = (pi) / (6), (3 pi) / (2), pi - (pi) / (6)$

$\Rightarrow x = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$ $=> x = (pi) / (6), (5 pi) / (6), (3 pi) / (2)$

How do you solve csc^2x-cscx+9=11csc2x−cscx+9=11csc^2x-cscx+9=11 for 0<=x<=2pi0≤x≤2π0<=x<=2pi?