How do I solve 1-sinx=3cosxfor [0, 2)? Thank you!

#1-sinx=3cosx# for #[0,2pi)#

1 Answer
Feb 9, 2018

Answer:

#x=pi/2,-0.9273..+2pi#

Explanation:

#1-sinx=3cosx#

#(1-sinx)^2=(3cosx)^2#

#1-2sinx+sin^2x=9cos^2x#

#1-2sinx+sin^2x=9(1-sin^2x)#

#1-2sinx+sin^2x=9-9sin^2x#

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

Let #u=sinx#:

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

#5u^2-u-4=0#

#(u-1)(5u+4)=0#

#u=1,-4/5#

Plug #sinx# back in for #u#:

#sinx=1, -4/5#

#x=pi/2,sin^-1(-4/5)~~-0.9273..#

To get the second number in the wanted range, just add #2pi#:

#x=pi/2,-0.9273..+2pi#