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}].
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Số câu hỏi: 40
VietJack
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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#
#color[red][[-1][-4]]# #=> color[red][YES]#
#[2][2]# #=> NO#
#[-2][-2]# #=> 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, //www.youtube.com/playlist?list=PLsX0tNIJwRTxqIIcfpsTII9_xhqVIbQlp