# Solve 5sin2x =4tanx

embizze | High School Teacher | (Level 1) Educator Emeritus

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)`

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

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The graph of the two functions:

sciencesolve | Teacher | (Level 3) Educator Emeritus

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.`