How do you solve 3cos^2x +5sinx - 1=03cos2x+5sinx1=0?

1 Answer
Mar 6, 2018

The solutions are x=sin^-1(-1/3), pi-sin^-1(-1/3)x=sin1(13),πsin1(13).

Explanation:

Use the Pythagorean identity to derive a slightly different one:

sin^2x+cos^2x=1sin2x+cos2x=1

=>cos^2x=1-sin^2xcos2x=1sin2x

Use this new identity to replace cos^2xcos2x in the actual problem, then solve it like a quadratic:

3color(red)(cos^2x)+5sinx-1=03cos2x+5sinx1=0

3(color(red)(1-sin^2x))+5sinx-1=03(1sin2x)+5sinx1=0

3-3sin^2x+5sinx-1=033sin2x+5sinx1=0

-3sin^2x+5sinx+2=03sin2x+5sinx+2=0

3sin^2x-5sinx-2=03sin2x5sinx2=0

3(sinx)^2-5sinx-2=03(sinx)25sinx2=0

Replace sinxsinx with uu:

3u^2-5u-2=03u25u2=0

(3u+1)(u-2)=0(3u+1)(u2)=0

u=-1/3, 2u=13,2

Substitute sinxsinx back in for uu:

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

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