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

2 Answers
Jun 4, 2018

arctan(1+sqrt(2)),arctan(1+sqrt(2))+pi,pi+arctan(1-sqrt(2)),2pi+arctan(1+sqrt(2))

Explanation:

Substituting
t=tan(theta)
so we get

t^2-2t-1=0
t_{1,2}=1pmsqrt(2)

Jun 4, 2018

67^@50; 157^@51; 247^@50: 337^@51

Explanation:

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