SOLVE to the nearest four decimals?

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SOLVE to the nearest four decimals?

$\sqrt{3} \cdot {(cot x)}^{2} + cot x = 1$ $sqrt(3)*(cotx)^2+cotx=1$

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Answer 1 · 2018-05-17T04:01:40

$x = 62^{\circ} 3393 + k 180^{\circ}$ $x = 62^@3393 + k180^@$
$3 x = - 42^{\circ} 2363 + k 180^{\circ}$ $3x = - 42^@2363 + k180^@$

Explanation:

Method 1
$\sqrt{3} (\frac{{cos}^{2} x}{{sin}^{2} x}) + \frac{cos x}{sin x} = 1$ $sqrt3(cos^2 x/(sin^2 x)) + cos x/(sin x) = 1$
$\sqrt{3} . {cos}^{2} x + sin x . cos x = {sin}^{2} x$ $sqrt3.cos^2 x + sin x.cos x = sin^2 x$
Divide both side by cos x, (condition cos x != 0)
$\sqrt{3} + tan x = {tan}^{2} x$ $sqrt3 + tan x = tan^2 x$
${tan}^{2} x - tan x - \sqrt{3} = 0$ $tan^2 x - tan x - sqrt3 = 0$
Solve this quadratic equation for tan x.
D = d^2 = b^2 - 4ac = 1 + 4sqrt3 = 7.9282 --> $d = \pm 2.8157$ $d = +- 2.8157$
There are 2 real roots:
$tan x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{1}{2} \pm \frac{2.8157}{2}$ $tan x = -b/(2a) +- d/(2a) = 1/2 +- 2.8157/2$
$tan x = 0.5 + 1.4079 = 1.9079$ $tan x = 0.5 + 1.4079 = 1.9079$
$tan x = 0.5 - 1.4079 = - 0.9079$ $tan x = 0.5 - 1.4079 = - 0.9079$
a. $tan x = 1.9079$ $tan x = 1.9079$
Calculator and unit circle give:
$x = 62^{\circ} 3393 + k 180^{\circ}$ $x = 62^@3393 + k180^@$
b. $tan x = - 0.9079$ $tan x = - 0.9079$
$x = - 42^{\circ} 2363 + k 180^{\circ}$ $x = -42^@2363 + k180^@$

Answer 2 · 2018-05-17T04:16:54

$x = 62^{\circ} 3405 + k 180^{\circ}$ $x = 62^@3405 + k180^@$
$x = - 42^{\circ} 2614 + k 180^{\circ}$ $x = -42^@2614 + k180^@$

Explanation:

Method 2
Call cot x = t, we get a quadratic equation to solve:
$\sqrt{3} t^{2} + t - 1 = 0$ $sqrt3t^2 + t - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 1 + 4 \sqrt{3} = 7.9282$ $D = d^2 = b^2 - 4ac = 1 + 4sqrt3 = 7.9282$ --> $d = \pm 2.8157$ $d = +- 2.8157$
There are 2 real roots:
$t = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{2 \sqrt{3}} \pm \frac{2.8157}{2 \sqrt{3}} =$ $t = -b/(2a) +- d/(2a) = - 1/(2sqrt3) +- 2.8157/(2sqrt3) =$
$= - 0.2887 \pm 0.8128$ $= - 0.2887 +- 0.8128$
cot x = t = 0.5241 --> $tan x = 1.9080$ $tan x = 1.9080$
cot x = t = - 1.1015 --> $tan x = - 0.9078$ $tan x = - 0.9078$
a. tan x = 1.9080
$x = 62^{\circ} 3405 + k 180^{\circ}$ $x = 62^@3405 + k180^@$
b. tan x = - 0.9087
$3 x = - 42^{\circ} 2614 + k 180^{\circ}$ $3x = - 42^@2614 + k180^@$

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SOLVE to the nearest four decimals?

$\sqrt{3} \cdot {(cot x)}^{2} + cot x = 1$ $sqrt(3)*(cotx)^2+cotx=1$

2 Answers

Explanation:

Explanation:

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SOLVE to the nearest four decimals?

√3⋅(cotx)2+cotx=1sqrt(3)*(cotx)^2+cotx=1

2 Answers

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$\sqrt{3} \cdot {(cot x)}^{2} + cot x = 1$ $sqrt(3)*(cotx)^2+cotx=1$