Question

Find all radian solutions on the interval `(-oo, oo)`? `sin(2x) - tanx = 0`

Answer 1

`x= pi/4+-pi/2n`
`x= pi+-2pin`
`x= 0+-2pin`
n is an element of all integers

Explanation:

`sin2x-tanx=0`

Apply sine double angle identity:
`2sinxcosx-tanx=0`

Factor out the `sinx`:
`sinx(2cosx-secx)=0`

Solve the 1st one:
`sinx=0`

`x=0, pi`

Solve the 2nd one:
`2cosx=secx`

`+-sqrt2=secx`

`1/(+-sqrt2)= 1/secx`

`+-sqrt2/2= cosx`

`x= pi/4, (3pi)/4, (5pi)/4, (7pi)/4`

General solutions:
`x= pi/4+-2pin`
`x= (3pi)/4+-2pin`
`x= (5pi)/4+-2pin`
`x= (7pi)/4+-2pin`
`x= pi+-2pin`
`x= 0+-2pin`

Where n is an element of all integers

NOTE:
`x= pi/4+-2pin`
`x= (3pi)/4+-2pin`
`x= (5pi)/4+-2pin`
`x= (7pi)/4+-2pin`

These four can be rewritten as:
`x= pi/4+-pi/2n`

Check with this graph:
graph{sin(2x)-tanx [-10, 10, -5, 5]}

Let me know if this makes sense, I tried it in another manner but lost solutions in the process of doing so.