三角函數問題

1.
tanθ+2cotθ=3
tanθ+2 / tanθ=3
(tanθ)^2 - 3 tanθ + 2 = 0
(tanθ - 1) (tanθ - 2) = 0

tanθ = 1 or 2

θ = arctan(1) + n兀, arctan (2) + n兀
for 0<=θ<=2兀
θ = 兀/4, 5兀/4, arctan(2) or 兀+arctan (2)

2.
sinθ=4cosθ
sinθ / cosθ = 4cosθ / cosθ
tanθ = 4

θ = arctan(4) + n兀
for 0<=θ<=2兀
θ = arctan(4) or 兀+arctan (4)