Find all values of Θ in the interval [0°, 360°)? tan²Θ - 2 tan Θ -1 = 0

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Find all values of Θ in the interval [0°, 360°)? tan²Θ - 2 tan Θ -1 = 0

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Answer 1 · 2018-06-04T14:21:22

$arctan (1 + \sqrt{2}), arctan (1 + \sqrt{2}) + π, π + arctan (1 - \sqrt{2}), 2 π + arctan (1 + \sqrt{2})$ $arctan(1+sqrt(2)),arctan(1+sqrt(2))+pi,pi+arctan(1-sqrt(2)),2pi+arctan(1+sqrt(2))$

Explanation:

Substituting
$t = tan (θ)$ $t=tan(theta)$
so we get

$t^{2} - 2 t - 1 = 0$ $t^2-2t-1=0$
$t_{1, 2} = 1 \pm \sqrt{2}$ $t_{1,2}=1pmsqrt(2)$

Answer 2 · 2018-06-04T15:53:18

$67^{\circ} 50; 157^{\circ} 51; 247^{\circ} 50 : 337^{\circ} 51$ $67^@50; 157^@51; 247^@50: 337^@51$

Explanation:

Solve this quadratic equation for tan t:
${tan}^{2} t - 2 tan t - 1 = 0$ $tan^2 t - 2tan t - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 4 + 4 = 8$ $D = d^2 = b^2 - 4ac = 4 + 4 = 8$ --> $d = \pm 2 \sqrt{2}$ $d = +- 2sqrt2$
The 2 real roots are:
$tan t = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{2}{2} \pm 2 \frac{\sqrt{2}}{2} = 1 \pm \sqrt{2}$ $tan t = -b/(2a) +- d/(2a) = 2/2 +- 2sqrt2/2 = 1 +- sqrt2$
a. $tan t = 1 + \sqrt{2} = 2.414$ $tan t = 1 + sqrt2 = 2.414$
Calculator and unit circle give 2 solutions for t:
$t = 67^{\circ} 50$ $t = 67^@50$ and $t = 67^{\circ} 50 + 180 = 247^{\circ} 50$ $t = 67^@50 + 180 = 247^@50$
b. $tan t = 1 - \sqrt{2} = - 0.414$ $tan t = 1 - sqrt2 = -0.414$
Calculator and unit circle give:
$t = - 22^{\circ} 49$ $t = - 22^@49$ and $t = - 22.49 + 180 = 157^{\circ} 51$ $t = - 22.49 + 180 = 157^@51$
Note: t = - 22.49 is co-terminal to t = 337.51
The answers for (0, 360) are:
$67^{\circ} 50; 157^{\circ} 51; 247^{\circ} 50; 337^{\circ} 51$ $67^@50; 157^@51; 247^@50; 337^@51$

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Find all values of Θ in the interval [0°, 360°)? tan²Θ - 2 tan Θ -1 = 0

2 Answers

Explanation:

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