Given: #-4sin^2(x)+10sin(2x)-1-4cos^2(x)=0#
Substitute, #-4sin^2(x)-4cos^2(x) =-4#:
#10sin(2x)-1-4=0#
Combine like termsL
#10sin(2x)-5=0#
Divide both sides by 10:
#sin(2x) -1/2 = 0#
Add #1/2# to both sides:
#sin(2x) = 1/2#
Using the inverse sine function on both sides:
#2x = sin^-1(1/2)#
#theta = sin^-1(1/2)# has two values within the range #0<= theta < 2pi#; they are, #pi/6#, and #(5pi)/6#:
#2x = pi/6# and #2x = (5pi)/6#
Add the integer multiples of #2pi#:
#2x = pi/6+ 2npi; n in ZZ# and #2x = (5pi)/6+ 2npi; n in ZZ#
Divide both sides of both equations by 2:
#x = pi/12+ npi; n in ZZ# and #x = (5pi)/12+ npi; n in ZZ#
