Use the Pythagorean identity to derive a slightly different one:
sin^2x+cos^2x=1sin2x+cos2x=1
=>cos^2x=1-sin^2x⇒cos2x=1−sin2x
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+5sinx−1=0
3(color(red)(1-sin^2x))+5sinx-1=03(1−sin2x)+5sinx−1=0
3-3sin^2x+5sinx-1=03−3sin2x+5sinx−1=0
-3sin^2x+5sinx+2=0−3sin2x+5sinx+2=0
3sin^2x-5sinx-2=03sin2x−5sinx−2=0
3(sinx)^2-5sinx-2=03(sinx)2−5sinx−2=0
Replace sinxsinx with uu:
3u^2-5u-2=03u2−5u−2=0
(3u+1)(u-2)=0(3u+1)(u−2)=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)