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Question Number 158523 by gsk2684 last updated on 05/Nov/21

find the number of values of p for which equation sin^3 x+1+p^3 −3p sin x =0(p>0) has a root?

findthenumberofvaluesofp forwhichequation sin3x+1+p3−3psinx=0(p>0) hasaroot?

Answered by mr W last updated on 05/Nov/21

let t=sin x −1≤t≤1 t^3 −3pt+p^3 +1=0 Δ=−p^3 +(((p^3 +1)/2))^2 =(((p^3 −1)/2))^2 ≥0 for p^3 >1, i.e. p>1: t=((((p^3 −1)/2)−((p^3 +1)/2)))^(1/3) −((((p^3 −1)/2)+((p^3 +1)/2)))^(1/3) t=−1−p −1≤−1−p≤1 −2≤p≤0 ≠ p>1 for p^3 <1, i.e. p<1: t=((((1−p^3 )/2)−((p^3 +1)/2)))^(1/3) −((((1−p^3 )/2)+((p^3 +1)/2)))^(1/3) t=−p−1 −1≤−p−1≤1 −2≤p≤0 ✓ for p^3 =1, i.e. p=1: t^3 −3t+2=0 t^3 −t^2 +t^2 −t−2t+2=0 (t−1)^2 (t+2)=0 ⇒t=1 ✓, t=−2 summary: −2≤p≤0 or p=1 for p>0 there is only one value p=1.

lett=sinx −1⩽t⩽1 t3−3pt+p3+1=0 Δ=−p3+(p3+12)2=(p3−12)2⩾0 forp3>1,i.e.p>1: t=p3−12−p3+123−p3−12+p3+123 t=−1−p −1⩽−1−p⩽1 −2⩽p⩽0≠p>1 forp3<1,i.e.p<1: t=1−p32−p3+123−1−p32+p3+123 t=−p−1 −1⩽−p−1⩽1 −2⩽p⩽0✓ forp3=1,i.e.p=1: t3−3t+2=0 t3−t2+t2−t−2t+2=0 (t−1)2(t+2)=0 ⇒t=1✓,t=−2 summary: −2⩽p⩽0orp=1 forp>0thereisonlyonevaluep=1.

Commented bygsk2684 last updated on 12/Nov/21

thank you sir (°⌣°)

thankyousir(°⌣°)

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