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Solve 5sin2x =4tanx

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user5451174 | eNotes Newbie

Posted May 12, 2013 at 5:27 PM via web

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Solve 5sin2x =4tanx

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

Posted May 12, 2013 at 7:04 PM (Answer #2)

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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)`

In radians `x~~.8861+npi"or"x~~2.555+npi`

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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:

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 12, 2013 at 6:15 PM (Answer #1)

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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.`

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