Given: #tan(x)+cot(x)=4sin(2x)#
Substitute #tan(x) = sin(x)/cos(x)# and #cot(x) = cos(x)/sin(x)#
#sin(x)/cos(x)+cos(x)/sin(x)=4sin(2x)#
Avoid any roots that would cause division by 0 by adding the following restrictions:
#sin(x)/cos(x)+cos(x)/sin(x)=4sin(2x), cos(x)!=0, sin(x)!=0#
We know that the two restrictions occur at integer multiples of #pi/2#
#sin(x)/cos(x)+cos(x)/sin(x)=4sin(2x), x!=npi/2, n in ZZ#
Substitute #sin(2x) = 2sin(x)cos(x)#:
#sin(x)/cos(x)+cos(x)/sin(x)=4(2sin(x)cos(x)), x!=npi/2, n in ZZ#
Eliminate the denominators by multiplying both sides by #sin(x)cos(x)#:
#sin^2(x)+cos^2(x)=4(2sin(x)cos(x))sin(x)cos(x), x!=npi/2, n in ZZ#
Substitute #sin^2(x)+cos^2(x) = 1#
#1=4(2sin(x)cos(x))sin(x)cos(x), x!=npi/2, n in ZZ#
Simplify the right side:
#1= 8sin^2(x)cos^2(x), x!=npi/2, n in ZZ#
Divide both sides by 2:
#4sin^2(x)cos^2(x) = 1/2, x!=npi/2, n in ZZ#
Write the left side as a single square:
#(2sin(x)cos(x))^2 = 1/2, x!=npi/2, n in ZZ#
Substitute #2sin(x)cos(x) = sin(2x)#
#sin^2(2x) = 1/2, x!=npi/2, n in ZZ#
#sin^2(2x) = 1/2, x!=npi/2, n in ZZ#
Take the square root of both sides:
#sin(2x) = +-sqrt(1/2), x!=npi/2, n in ZZ#
Take the inverse sine of both sides:
#2x = sin^-1(-sqrt(1/2))# and #2x=sin^-1(sqrt(1/2)), x!=npi/2, n in ZZ#
#2x = -pi/4# and #2x=pi/4, x!=npi/2, n in ZZ#
We know that this repeats at integer multiples of #pi# and it will not violate the domain restrictions:
#2x = kpi -pi/4# and #2x= kpi + pi/4, k in ZZ#
Divide both equations by 2:
#x = kpi/2 -pi/8# and #x= kpi/2 + pi/8, k in ZZ#