Toán Lớp 11: Sin^2 3x=4cos4x+3 giúp mình bài nay cảm ơn mn

Question

Toán Lớp 11:  Sin^2 3x=4cos4x+3 giúp mình bài nay cảm ơn mn

mocmien87 · Answer

sin^2 3x=4cos4x+3$ $<=>(1-cos6x)/2=8cos^2 2x-4+3$ $<=>1-cos(3.2x)=16cos^2 2x-2$ $<=>1-4cos^3 2x+3cos2x=16cos^2 2x-2$ $<=>4cos^3 2x+16cos^2 2x-3cos2x-3=0$ $<=> ( left[ begin{array}{l}cos2x= dfrac{1}{2} cos2x= dfrac{-9+ sqrt{57}}{4} end{array} right. ) $ $<=> ( left[ begin{array}{l}x=± dfrac{π}{6}+kπ x= dfrac{±arccos( dfrac{-9+ sqrt{57}}{4})}{2}+kπ end{array} right. )

buianhtuan · Answer

$ begin{array}{l} { sin ^2}3x = 4 cos 4x + 3    dfrac{{1 -  cos 6x}}{2} = 4 cos 4x + 3     Leftrightarrow 1 -  cos 6x = 8 cos 4x + 6     Leftrightarrow 8 cos 4x +  cos 6x =  - 5     Leftrightarrow 8 left( {2{{ cos }^2}2x - 1}  right) + 4{ cos ^3}2x - 3 cos 2x =  - 5     Leftrightarrow 4{ cos ^3}2x + 16{ cos ^2}2x - 3 cos 2x - 3 = 0     Leftrightarrow 4{ cos ^3}2x - 2{ cos ^2}2x + 18{ cos ^2}2x - 9 cos 2x + 6 cos 2x - 3 = 0     Leftrightarrow 2{ cos ^2}2x left( {2 cos 2x - 1}  right) + 9 cos 2x left( {2 cos 2x - 1}  right) + 3 left( {2 cos 2x - 1}  right) = 0     Leftrightarrow  left( {2 cos 2x - 1}  right) left( {2{{ cos }^2}2x + 9 cos 2x + 3}  right) = 0     Leftrightarrow  left[  begin{array}{l}  cos 2x =  dfrac{1}{2}    cos 2x =  dfrac{{ - 9 -  sqrt {57} }}{4}  end{array}  right.     Leftrightarrow  left[  begin{array}{l} 2x =   pm  dfrac{ pi }{3} + k2 pi    2x =   pm  arccos  left( { dfrac{{ - 9 + sqrt {57} }}{4}}  right) + k2 pi   end{array}  right.     Leftrightarrow  left[  begin{array}{l} x =   pm  dfrac{ pi }{6} + k pi    x =  dfrac{{  pm  arccos  left( { dfrac{{ - 9 + sqrt {57} }}{4}}  right)}}{2} + k pi   end{array}  right. left( {k  in  mathbb{Z}}  right)  end{array}$