How do you solve $2 cot x - \frac{4}{cot x} = 7$ $2cotx-4/cotx=7$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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How do you solve $2 cot x - \frac{4}{cot x} = 7$ $2cotx-4/cotx=7$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2017-06-14T03:44:12

$14^{\circ} 04; - 63^{\circ} 43$ $14^@04; -63^@43$

Explanation:

$2 cot x - \frac{4}{cot x} - 7 = 0$ $2cot x - 4/(cot x) - 7 = 0$
$2 {cot}^{2} x - 7 cot x - 4 = 0$ $2cot^2 x - 7cot x - 4 = 0$
Solve this quadratic equation for cot x.
$D = d^{2} = b^{2} - 4 a c = 49 + 32 = 81$ $D = d^2 = b^2 - 4ac = 49 + 32 = 81$ --> $d = \pm 9$ $d = +- 9$
There are 2 real roots:
$cot x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{7}{4} \pm \frac{9}{4}$ $cot x = -b/(2a) +- d/(2a) = 7/4 +- 9/4$
a. $cot x = \frac{16}{4} = 4$ $cot x = 16/4 = 4$ --> $tan x = \frac{1}{4}$ $tan x = 1/4$
Calculator gives -->
$x = 14^{\circ} 04$ $x = 14^@04$
b. $cot x = - \frac{2}{4} = - \frac{1}{2}$ $cot x = -2/4 = -1/2$ --> $tan x = - 2$ $tan x = - 2$
$x = - 63^{\circ} 43$ $x = -63^@43$

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How do you solve $2 cot x - \frac{4}{cot x} = 7$ $2cotx-4/cotx=7$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Explanation:

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How do you solve 2cotx-4/cotx=72cotx−4cotx=72cotx-4/cotx=7 in the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?

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How do you solve $2 cot x - \frac{4}{cot x} = 7$ $2cotx-4/cotx=7$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?