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sin-4x-sin-3x-sin-2x-find-x-




Question Number 163822 by mkam last updated on 11/Jan/22
sin(4x) − sin(3x) = sin(2x) find x ?
sin(4x)sin(3x)=sin(2x)findx?
Answered by cortano1 last updated on 11/Jan/22
Commented by mr W last updated on 11/Jan/22
please recheck. i think answer is  wrong.
pleaserecheck.ithinkansweriswrong.
Commented by cortano1 last updated on 11/Jan/22
what wrong !
whatwrong!
Commented by mr W last updated on 11/Jan/22
x=±cos^(−1) ((2−((10))^(1/3) −((100))^(1/3) )/6)+2kπ doesn′t  fulfill the equation.
x=±cos1210310036+2kπdoesntfulfilltheequation.
Answered by mr W last updated on 11/Jan/22
sin 4x−sin 2x=sin 3x  sin (3x+x)−sin (3x−x)=sin 3x  2 cos 3x sin x=sin 3x  2(4 cos^3  x−3 cos x)sin x=sin x (3−4 sin^2  x)  ⇒sin x=0 ⇒x=nπ  or  8 cos^3  x−6 cos x=3−4 sin^2  x  8 cos^3  x−4 cos^2  x−6 cos x+1=0  cos^3  x−(1/2) cos^2  x−(3/4) cos x+(1/8)=0  let cos x=t+(1/6)  t^3 −((5t)/6)−(1/(108))=0  t=((√(10))/3) sin (((2kπ)/3)−(1/3)sin^(−1) ((√(10))/(100)))  cos x=(1/6)+((√(10))/3) sin (((2kπ)/3)−(1/3)sin^(−1) ((√(10))/(100)))  such that −1≤cos x≤1, ⇒ k=0, 2  cos x=(1/6)−((√(10))/3) sin ((1/3)sin^(−1) ((√(10))/(100))) >0  cos x=(1/6)−((√(10))/3) sin ((π/3)−(1/3)sin^(−1) ((√(10))/(100))) <0  ⇒x=2nπ±cos^(−1) {(1/6)−((√(10))/3) sin ((1/3)sin^(−1) ((√(10))/(100)))}  ⇒x=(2n+1)π±cos^(−1) {((√(10))/3) sin ((π/3)−(1/3)sin^(−1) ((√(10))/(100)))−(1/6)}
sin4xsin2x=sin3xsin(3x+x)sin(3xx)=sin3x2cos3xsinx=sin3x2(4cos3x3cosx)sinx=sinx(34sin2x)sinx=0x=nπor8cos3x6cosx=34sin2x8cos3x4cos2x6cosx+1=0cos3x12cos2x34cosx+18=0letcosx=t+16t35t61108=0t=103sin(2kπ313sin110100)cosx=16+103sin(2kπ313sin110100)suchthat1cosx1,k=0,2cosx=16103sin(13sin110100)>0cosx=16103sin(π313sin110100)<0x=2nπ±cos1{16103sin(13sin110100)}x=(2n+1)π±cos1{103sin(π313sin110100)16}
Commented by peter frank last updated on 11/Jan/22
thank you
thankyou

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