0<λ<1
3sinx-4sin3x-λ=0
sin3x=λ=sinα
3x=λ=π-λ
x=λ/3,π/3-λ/3
A+B+C=Ï€
λ/3+π/3-λ/3+C=π
C=Ï€-Ï€/3=2Ï€/3
In a triangle ABC, angleA is greater than angleB and A,B,C are the roots of the equation 3sinx-4sin3x-λ=0.Find the angleC.(0<λ<1)
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3 Answers
Manish Shankar
·2013-06-14 01:13:00
is it 3sinx-4sin3x-λ=0
or 3sinx-4sin3x-λ=0
- Anomitro Kundu oh sorry sir it is 3sinx-4sin cubexUpvote·0· Reply ·2013-06-14 01:17:48
Anomitro Kundu
·2013-06-14 01:15:49