How do I solve #5 cos 2x-6sin x-1=0# for #(0,2pi)#?
1 Answer
Nov 21, 2017
Explanation:
#"using the "color(blue)"trigonometric identity"#
#•color(white)(x)cos2x=1-2sin^2x#
#rArr5(1-2sin^2x)-6sinx-1=0#
#rArr5-10sin^2x-6sinx-1=0#
#rArr-10sin^2x-6sinx+4=0#
#rArr-2(5sin^2x+3sinx-2)=0#
#"we have a quadratic in sine"#
#rArr-2(5sinx-2)(sinx+1)=0#
#rArr5sinx-2=0" or "sinx+1=0#
#•color(white)(x)sinx=2/5#
#rArrx=0.41" or "x=(pi-0.41)=2.73to("2 dec. places")#
#•color(white)(x)sinx=-1rArrx=(3pi)/2#
#rArrx=0.41,2.73,(3pi)/2;x in(0,2pi)#