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#