Question #6e0d6

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Answer 1 · 2018-02-19T17:50:28

#x# is about #pi/6, (5pi)/6,0.8481,2.294#.

#4cos2x+10sinx-7=0#

#4(1-2sin^2x)+10sinx-7=0#

#4-8sin^2x+10sinx-7=0#

#-8sin^2x+10sinx-3=0#

#8sin^2x-10sinx+3=0#

Let #u = sinx#:

#8u^2-10u+3=0#

#(2u-1)(4s-3)=0#

#u=1/2,3/4#

Plug #sinx# back in for #u#:

#sinx=1/2,sinx=3/4#

#=>x=sin^-1(1/2),sin^-1(3/4)# and the supplementary angles to each of these.

#=>x=pi/6, pi-pi/6, sin^-1(3/4), pi-sin^-1(3/4)#

#color(white)(=>x)~~pi/6, (5pi)/6,0.8481,2.294#