# Solve 5sin2x =4tanx

Posted on

Solve `5sin(2x)=4tan(x)`

Use `sin(2x)=2sin(x)cos(x)` and `tan(x)=(sin(x))/(cos(x))` to get

`10sinxcosx=4sinx/cosx`

You cannot divide both sides by sinx -- you might lose a root. (In this case sinx=0 is a solution since both sides of the equation will be 0. If you recognize this you can divide and continue with the realization that sinx=0 is a solution.) It is better to proceed:

`10sinxcosx-4sinx/cosx=0`

`2sinx(5cosx-2/cosx)=0`

By the zero product property :

`2sinx=0==>sinx=0==>x=npi`

or

`5cosx-2/cosx=0`

`==>5cosx=2/cosx`

`==>cos^2x=2/5`

`==>cosx=+-sqrt(2/5)`

`==>x=cos^(-1)sqrt(2/5)"or"cos^(-1)-sqrt(2/5)`

----------------------------------------------------------------

The complete solution is `x=npi,cos^(-1)sqrt(2/5)+npi,cos^(-1)-sqrt(2/5)+npi`

---------------------------------------------------------------

The graph of the two functions:

Posted on

Since the request of the problem is vague, since it does not specify the argument of tangent function, you may solve the problem considering the argument x,such that:

`5sin 2x = 4tan x`

Using the double angle formula yields:

`sin 2x = 2 sin x cos x`

`5*2 sin x cos x = 4 tan x`

Using the following trigonometric identity for `tan x,` yields:

`tan x = sin x/cos x`

`10 sin x*cos x = 4sin x/cos x`

Reducing duplicate factors yields:

`10 cos x = 4/cos x => 10cos^2 x = 4 => cos^2 x = 4/10`

`cos x^2 = 2/5 => cos x = +-sqrt(2/5)`

`x = cos^(-1)(+-sqrt(2/5)) + 2npi`

Hence, evaluating the solutions to the given equation, considering the argument of tangent function as x, yields `x = +-cos^(-1)(+-sqrt(2/5)) + 2npi.`