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Question Number 155244 by mathdanisur last updated on 27/Sep/21

Solve the equation:    (sinx)^3  + sinx = cosx

Solvetheequation:(sinx)3+sinx=cosx

Answered by ajfour last updated on 27/Sep/21

tan x((1/(1+(1/(tan^2 x))))+1)=1  m(2m^2 +1)=1+m^2   2m^3 −m^2 +m−1=0  x=tan^(−1) m+kπ

tanx(11+1tan2x+1)=1m(2m2+1)=1+m22m3m2+m1=0x=tan1m+kπ

Commented by mathdanisur last updated on 27/Sep/21

very nice Ser, thank you

veryniceSer,thankyou

Commented by mathdanisur last updated on 27/Sep/21

Thanks Ser, but m=?

ThanksSer,butm=?

Commented by mr W last updated on 27/Sep/21

m^3 −(m^2 /2)+(m/2)−(1/2)=0  m=n+(1/6)  n^3 +((5n)/(12))−((23)/(54))=0  m=tan x=(1/6)+((((√(249))/(72))+((23)/(108))))^(1/3) −((((√(249))/(72))−((23)/(108))))^(1/3) ≈0.739  x=kπ+tan^(−1) ((1/6)+((((√(249))/(72))+((23)/(108))))^(1/3) −((((√(249))/(72))−((23)/(108))))^(1/3) )

m3m22+m212=0m=n+16n3+5n122354=0m=tanx=16+24972+231083249722310830.739x=kπ+tan1(16+24972+23108324972231083)

Answered by MJS_new last updated on 27/Sep/21

sin^3  x +sin x =cos x  sin^3  x +sin x =±(√(1−sin^2  x))  sin x =s  s^3 +s=±(√(1+s^2 ))  squaring [might introduce false solutions]  and transforming  s^6 +2s^4 +2s^2 −1=0  (s^2 )^3 +2(s^2 )^2 +2(s^2 )=0  s^2 =u−(2/3)  u^3 +(2/3)u−((47)/(27))=0  Cardano′s solution  u=((((47)/(54))+((√(249))/(18))))^(1/3) −((−((47)/(54))+((√(249))/(18))))^(1/3)   it makes no sense to go on with this  u≈1.01987663  ⇒  s^2 =.353209964  ⇒  sin x ≈±.594314701  testing leads to  sin x ≈.594314701  ⇒  x≈2nπ+2.50517939∨x≈2nπ+.636413265

sin3x+sinx=cosxsin3x+sinx=±1sin2xsinx=ss3+s=±1+s2squaring[mightintroducefalsesolutions]andtransformings6+2s4+2s21=0(s2)3+2(s2)2+2(s2)=0s2=u23u3+23u4727=0Cardanossolutionu=4754+2491834754+249183itmakesnosensetogoonwiththisu1.01987663s2=.353209964sinx±.594314701testingleadstosinx.594314701x2nπ+2.50517939x2nπ+.636413265

Commented by mathdanisur last updated on 27/Sep/21

Very nice Ser, thank you

VeryniceSer,thankyou

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