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, https://www.youtube.com/playlist?list=PLsX0tNIJwRTxqIIcfpsTII9_xhqVIbQlp
