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Question Number 210231 by a.lgnaoui last updated on 03/Aug/24

Resoudre dans R   { ((acos x−bsin x=c     (x≠0))),((sin ((1/(sin x)))         =d    (−1≤d≤+1))) :}

ResoudredansR{acosxbsinx=c(x0)sin(1sinx)=d(1d+1)

Commented by mr W last updated on 04/Aug/24

they are two different equations for  variable x. they have different roots.  they can not be an equation system!    eqn. 1:  a cos x−b sin x=c  (√(a^2 +b^2 )) cos (x+tan^(−1) (b/a))=c  assume ∣c∣≤(√(a^2 +b^2 ))  cos (x+tan^(−1) (b/a))=(c/( (√(a^2 +b^2 ))))  ⇒x+tan^(−1) (b/a)=2kπ±cos^(−1) (c/( (√(a^2 +b^2 ))))  ⇒x=2kπ−tan^(−1) (b/a)±cos^(−1) (c/( (√(a^2 +b^2 ))))  eqn. 2:  sin ((1/(sin x)))=d  ⇒(1/(sin x))=kπ+(−1)^k  sin^(−1) d  ⇒sin x=(1/(kπ+(−1)^k  sin^(−1) d))  ⇒x=nπ+(−1)^n sin^(−1) ((1/(kπ+(−1)^k  sin^(−1) d)))

theyaretwodifferentequationsforvariablex.theyhavedifferentroots.theycannotbeanequationsystem!eqn.1:acosxbsinx=ca2+b2cos(x+tan1ba)=cassumec∣⩽a2+b2cos(x+tan1ba)=ca2+b2x+tan1ba=2kπ±cos1ca2+b2x=2kπtan1ba±cos1ca2+b2eqn.2:sin(1sinx)=d1sinx=kπ+(1)ksin1dsinx=1kπ+(1)ksin1dx=nπ+(1)nsin1(1kπ+(1)ksin1d)

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