Question

How do I solve: `tan(x)+cot(x)=4sin(2x)` using double angle or half angle identities?

Answer 1

`x=k*pi/2+(-1)^k*(+-pi/8), k in ZZ`.

Explanation:

`tanx+cotx=4sin2x`.

`:. sinx/cosx+cosx/sinx=4sin2x`.

`:. (sin^2x+cos^2x)/(sinxcosx)=4sin2x`.

`:.1/(sinxcosx)xx2/2=4sin2x`.

`:.2/(2sinxcosx)=4sin2x`.

`:.2/(sin2x)=4sin2x`.

`:. 2/4=1/2=sin^2 2x`.

`:. sin2x=+-1/sqrt2=sin(+-pi/4)`.

Since, `sinu=sinv rArr u=kpi+(-1)^k*v, k in ZZ`, we have,

`2x=kpi+(-1)^k*(pmpi/4), k in ZZ`, or,

`x=k*pi/2+(-1)^k*(+-pi/8), k in ZZ`, is the desired solution.

Answer 2

`x = kpi/2 -pi/8` and `x= kpi/2 + pi/8, k in ZZ`

Explanation:

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`