How do you solve #5sin^2x-4sinx-1=0# in the interval #0<=x<=2pi#?

1 Answer
Nov 18, 2016

#" "x=pi/2" "Or " "x=-0.201358#

Explanation:

Let #color(blue)(y=sinx)#
#" "#
#5sin^2 x - 4sinx = 1 =0#
#" "#
#rArr5y^2 - 4y - 1 =0#
#" "#
This equation is solved by factorizing #" "5y^2 - 4y - 1" "# and
#" "#
finding its quadratic roots.
#" "#
#delta = b^2 - 4ac = (-4)^2-4(5)(-1) = 16 + 20 = 36#
#" "#
#delta>0" "# Two roots exist in #RR#
#" "#
#y_1=(-b- sqrt(delta))/(2a)=(4-sqrt36)/(2xx5)=(4-6)/10=-2/10=-1/5#
#" "#
#y_2=(-b + sqrt delta)/(2a)=(4+sqrt36)/(2xx5)=(4+6)/10=10/10=1#
#" "#
The form of the equation by substituting the factorization form of the polynomial :
#" "#
#5y^2 - 4y - 1=0#
#" "#
#(y+1/5)(y-1)=0#
#" "#
#y+1/5=0rArry=-1/5rArrcolor(blue)(sinx=-1/5)rArrx=-0.2#
#" "#
#y-1=0rArry=1rArrcolor(blue)(sinx=1)rArrx=pi/2#
#" "#
Hence,#" "x=pi/2" "Or " "x=-0.201358#