Nghiệm của phương trình 2sin^2x-5sinx+2=0
|
Giải phương trình: (2{sin ^2}x - 5sin x + 2 = 0) A. (S = left{ {dfrac{pi }{6} + k2pi ;,,dfrac{{5pi }}{6} + k2pi |k in Z} right}). B. (S = left{ {dfrac{pi }{6} + k2pi ;,,-dfrac{{5pi }}{6} + k2pi |k in Z} right}). C. (S = left{ {-dfrac{pi }{6} + k2pi ;,,dfrac{{5pi }}{6} + k2pi |k in Z} right}). D. (S = left{ {-dfrac{pi }{6} + k2pi ;,,-dfrac{{5pi }}{6} + k2pi |k in Z} right}).
Câu hỏi trên thuộc đề trắc nghiệm dưới đây ! Số câu hỏi: 40
Bằng cách đăng ký, bạn đồng ý với Điều khoản sử dụng và Chính sách Bảo mật của chúng tôi.
Giải phương trình: \(2{\sin ^2}x - 5\sin x + 2 = 0\)
A. \(S = \left\{ {\dfrac{\pi }{6} + k2\pi ;\,\,\dfrac{{5\pi }}{6} + k2\pi |k \in Z} \right\}\). B. \(S = \left\{ {\dfrac{\pi }{6} + k2\pi ;\,\,-\dfrac{{5\pi }}{6} + k2\pi |k \in Z} \right\}\). C. \(S = \left\{ {-\dfrac{\pi }{6} + k2\pi ;\,\,\dfrac{{5\pi }}{6} + k2\pi |k \in Z} \right\}\). D. \(S = \left\{ {-\dfrac{\pi }{6} + k2\pi ;\,\,-\dfrac{{5\pi }}{6} + k2\pi |k \in Z} \right\}\). Use the substitution method and then factor Let #u=sinx# so, #u^2=sin^2x# #2color(red)(u^2)-5color(red)(u)+2=0# Now you can factor Multiply the coefficient of the first term, #2#, with the last term, #2#. #2*2=4# Ask yourself what are the factors of #4# that add up to the coefficient of the middle term, #-5#? Factors of 4 #(1)(4)# #=> NO# Now place those factors in an order that makes it easy to factor by grouping. #(2u^2color(red)(-4u))+(color(red)(-1u)+2)=0# Factor out #2u# from the first grouping. Factor out #-1# from the second grouping. #color(red)(2u)(u-2)color(red)(-1)(u-2)=0# Now you can factor out a grouping, #(u-2)# #(u-2)(2u-1)=0# Now use the Zero Property #u-2=0# and #2u-1=0# #u=2# and #u=1/2# Now switch back to #sinx# #sin x=2# and #sin x =1/2# #cancel(sin x=2)# is discarded because #sin x# oscillates between -1 and 1. Falling back to trigonometry #sin x# is in the ratio #y/r# #y/r=1/2# which means that #x=sqrt3# by using the pythagorean theorem #x=sqrt(2^2-1^2)=sqrt(4-1)=sqrt3# or knowing that we have the special triangle , #30,60,90# which corresponds to the sides #1,sqrt3,2.# The side of length #1# corresponds to #30# degrees which is also #pi/6# All of that to say that #sin (pi/6)=1/2# #x={pi/6}# I have tutorials on methods of factoring found here, https://www.youtube.com/playlist?list=PLsX0tNIJwRTxqIIcfpsTII9_xhqVIbQlp |
