f.4 trigonometry!_!

(1) sec2 θ + tan θ = 3

1 + tan2 θ + tan θ = 3

tan2 θ + tan θ - 2 = 0

(tan θ + 2)(tan θ - 1) = 0

tan θ = 1 or -2

θ = π/4, 5π/4, 2.03 or 5.18

(2) √ 2 tan θ = secθ + 1

2 tan2 θ = sec2 θ + 2sec θ + 1

2sec2 θ - 2 = sec2 θ + 2sec θ + 1

sec2 θ - 2sec θ - 3 = 0

(sec θ - 3)(sec θ + 1) = 0

sec θ = 3 or -1

θ = π or 1.23

(3) sec θ - 4tan θ + 7 = 0

sec θ = 4tan θ - 7

sec2 θ = 16tan2 θ - 56tan θ + 49

1 + tan2 θ = 16tan2 θ - 56tan θ + 49

15tan2 θ - 56tan θ + 48 = 0

(3tan θ - 4)(5tan θ - 12) = 0

tan θ = 4/3 or 12/5

θ = 0.927 or 1.176


2010-01-06 22:59:33 補充：
(3) Since squaring method is used, we have to check back those ans by substituting them into the original equation.

After checking, only θ = 1.176 is the true ans.