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x-1-and-x-2-are-solutions-of-equality-cos-pix-pi-6-sin-pix-pi-6-1-2-3-0-x-12-Find-the-value-of-x-1-x-2-




Question Number 143964 by naka3546 last updated on 20/Jun/21
x_1  and  x_2   are  solutions  of  equality :    cos (((πx+π)/6)) − sin (((πx−π)/6)) = (1/2) (√3)   ,   0 ≤ x ≤ 12  Find  the  value  of  x_1 + x_2  .
x1andx2aresolutionsofequality:cos(πx+π6)sin(πxπ6)=123,0x12Findthevalueofx1+x2.
Commented by Canebulok last updated on 20/Jun/21
   Solution:  By squaring both sides,  ⇒ cos(((πx)/6) + (π/6))−sin(((πx)/(6 )) − (π/6)) = ((√3)/2)  ⇒ [cos(((πx)/6))cos((π/6))−sin(((πx)/6))sin((π/6))] − [sin(((πx)/6))cos((π/6))−cos(((πx)/6))sin((π/6))] = ((√3)/2)
Solution:Bysquaringbothsides,cos(πx6+π6)sin(πx6π6)=32[cos(πx6)cos(π6)sin(πx6)sin(π6)][sin(πx6)cos(π6)cos(πx6)sin(π6)]=32
Commented by Canebulok last updated on 20/Jun/21
Solution:  ⇒ [cos(((πx)/6))cos((π/6))−sin(((πx)/6))cos((π/6))] + [cos(((πx)/6))sin((π/6))−sin(((πx)/6))sin((π/6))] = ((√3)/2)  ⇒ cos((π/6))[cos(((πx)/6))−sin(((πx)/6))] + sin((π/6))[cos(((πx)/6))−sin(((πx)/6))] = ((√3)/2)  ⇒ [cos((π/6))+sin((π/6))][cos(((πx)/6))−sin(((πx)/6))] = ((√3)/2)     By squaring both sides,  ⇒ [1+2cos((π/6))sin((π/6))][1−2sin(((πx)/6))cos(((πx)/6))] = (3/4)  ⇒ [1+sin((π/3))][1−sin(((πx)/3))] = (3/4)  ⇒ 1−sin(((πx)/3)) = (3/(4[1+sin((π/3))]))     ⇒ −sin(((πx)/3)) = (3/((4+4sin((π/3))))) − 1     ⇒ sin(((πx)/3)) = 1 − (3/((4+4sin((π/3)))))  ⇒ arcsin[1−(3/((4+4sin((π/3)))))] = ((πx)/3)  ⇒ arcsin[1−(3/((4+4sin((π/3)))))] ((3/π)) = x     ∼ Kevin
Solution:[cos(πx6)cos(π6)sin(πx6)cos(π6)]+[cos(πx6)sin(π6)sin(πx6)sin(π6)]=32cos(π6)[cos(πx6)sin(πx6)]+sin(π6)[cos(πx6)sin(πx6)]=32[cos(π6)+sin(π6)][cos(πx6)sin(πx6)]=32Bysquaringbothsides,[1+2cos(π6)sin(π6)][12sin(πx6)cos(πx6)]=34[1+sin(π3)][1sin(πx3)]=341sin(πx3)=34[1+sin(π3)]sin(πx3)=3(4+4sin(π3))1sin(πx3)=13(4+4sin(π3))arcsin[13(4+4sin(π3))]=πx3arcsin[13(4+4sin(π3))](3π)=xKevin
Answered by bramlexs22 last updated on 20/Jun/21
sin (((πx−π)/6))=cos (((3π−πx+π)/6))=cos (((πx−4π)/6))  ⇔ cos (((πx+π)/6))−cos (((πx−4π)/6))=((√3)/2)  ⇔−2sin (((2πx−3π)/(12)))sin (((5π)/(12)))=((√3)/2)  ⇔
sin(πxπ6)=cos(3ππx+π6)=cos(πx4π6)cos(πx+π6)cos(πx4π6)=322sin(2πx3π12)sin(5π12)=32

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