Question

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

Answer 1

`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)`

Answer 2

`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`