How do you solve #3cos^2x +5sinx - 1=0#?

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Answer 1 · 2018-03-06T04:12:50

The solutions are #x=sin^-1(-1/3), pi-sin^-1(-1/3)#.

Use the Pythagorean identity to derive a slightly different one:

#sin^2x+cos^2x=1#

#=>cos^2x=1-sin^2x#

Use this new identity to replace #cos^2x# in the actual problem, then solve it like a quadratic:

#3color(red)(cos^2x)+5sinx-1=0#

#3(color(red)(1-sin^2x))+5sinx-1=0#

#3-3sin^2x+5sinx-1=0#

#-3sin^2x+5sinx+2=0#

#3sin^2x-5sinx-2=0#

#3(sinx)^2-5sinx-2=0#

Replace #sinx# with #u#:

#3u^2-5u-2=0#

#(3u+1)(u-2)=0#

#u=-1/3, 2#

Substitute #sinx# back in for #u#:

#sinx=-1/3,color(red)cancelcolor(black)(sinx=2)#

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