Find the general solution of the equation
Sin3θ = cosθ.
Hence, or otherwise, calculate the values of θ satisfying the given equation and 0 ＜θ＜π/2.
答案：(nπ)/2 + π/8, nπ + π/4; π/8, π/4
※ 請列明步驟！※
一條關於Trigonometry的簡單A.Maths.問題
2008-12-25 11:26 pm
回答 (2)
2008-12-26 12:42 am
2008-12-26 1:15 am
Sin3θ = cosθ
cosθ- sin(3θ) = 0
cosθ- cos(90 - 3θ) = 0
sin(45-θ)sin(-45+2θ)=0
sin(45-θ)=0 or sin(-45+2θ)=0
π/4-θ=nπ or -π/4+2θ=nπ
θ=nπ+π/4 or θ=nπ/2+π/8
the values of θ satisfying the given equation and 0 ＜θ＜π/2. are π/8, π/4
cosθ- sin(3θ) = 0
cosθ- cos(90 - 3θ) = 0
sin(45-θ)sin(-45+2θ)=0
sin(45-θ)=0 or sin(-45+2θ)=0
π/4-θ=nπ or -π/4+2θ=nπ
θ=nπ+π/4 or θ=nπ/2+π/8
the values of θ satisfying the given equation and 0 ＜θ＜π/2. are π/8, π/4
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