Trig help please..1 Question:)?

one way

replace tan theta with A

A² - 3A + 2 = 0

factor

(A - 1)(A - 2) = 0 ...yeah, yeah, I know it says quad formula, but this is way quicker.

A = 1 or A = 2

replace A back with tanΘ

tanΘ = 1 or tanΘ = 2

so Θ=π/4 + nπ/2 for all integers n, or Θ≈1.107 + nπ/2 for all integers n

(answers in radians)

Let u = tan(theta) to get u^2 - 3u + 2 = 0

By the quadratic formula, u = (3 +- sqrt(9 - 4*1*2))/2

u = (3 + 1)/2 = 2 or u = (3 - 1)/2 = 1

Then; 2 = tan (theta) or 1 = tan (theta)

let u = tan theta so the equation becomes

u^2 - 3u + 2 = 0

(- b +- sqrt(b^2 - 4ac)/2a

a = 1

b = - 3

c = 2

u = (3 +- sqrt((- 3)^2 - 4(1)(2))/((2)(1))

u = (3 =-sqrt(9 - 8))/2

u = (3 +- sqrt(1))/2

u = (3 +- 1)/2

u = 4/2 = 2 or 2/2 = 1

but we want to know what tan theta equals so substitute

set tan theta = 2 and solve and we get 63.43

and tan theta = 1 and we get 45

let's say that tan theta is equivalent to t

so tan^2 thta -tan theta +2=0 becomes t^2 - 3t +2=0

delta=1 so the roots are t'=1 and t''=2

and you solve tan theta =1 so theta = 45 degrees or tan theta=2 so thta = 60 degrees

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