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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?
findthenumberofvaluesofpforwhichequationsin3x+1+p33psinx=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=sinx1t1t33pt+p3+1=0Δ=p3+(p3+12)2=(p312)20forp3>1,i.e.p>1:t=p312p3+123p312+p3+123t=1p11p12p0p>1forp3<1,i.e.p<1:t=1p32p3+1231p32+p3+123t=p11p112p0forp3=1,i.e.p=1:t33t+2=0t3t2+t2t2t+2=0(t1)2(t+2)=0t=1,t=2summary:2p0orp=1forp>0thereisonlyonevaluep=1.
Commented by gsk2684 last updated on 12/Nov/21
thank you sir (°⌣°)
thankyousir(°°)

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