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Solve the equation `4 sin^2 x+12sinx cosx-cos^2x+5=0` for 0```<=` x`<=` 2`pi` ` `

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roshan-rox | Valedictorian

Posted June 4, 2013 at 4:35 PM via web

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Solve the equation `4 sin^2 x+12sinx cosx-cos^2x+5=0`

for 0```<=` x`<=` 2`pi`

` `

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

Posted June 4, 2013 at 5:07 PM (Answer #1)

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Solve `4sin^2x+12sinxcosx-cos^2x+5=0`

Rewrite the left hand side:

`4sin^2x+12sinxcosx-(1-sin^2x)+5=0`

`5sin^2x+12sinxcosx+4=0`

`5sin^2x+12sinxcosx+4(sin^2x+cos^2x)=0`

`9sin^2x+12sinxcosx+4cos^2x=0`

`(3sinx+2cosx)^2=0`  take the square root of both sides:

`3sinx+2cosx=0`

`3sinx=-2cosx`  Since sinx and cosx share no zeros, we can divide:

`tanx=-2/3`

`x="Tan"^(-1)(-2/3)`

With the restriction `0<=x<=2pi` we have:

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`x~~2.55359005,5.695182704`

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Note that tangent is negative in the second and fourth quadrants.

The graph:

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 4, 2013 at 5:15 PM (Answer #2)

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`4sin^2x+12sinxcosx-cos^2x+5 = 0`

We know the following.

`cos2x = 1-2sin^2x rarr sin^2x = (1-cos2x)/2`

`cos2x = 2cos^2x-1 rarr cos^2x = (1+cos2x)/2`

`sin2x = 2sinxcosx`

 

`4sin^2x+12sinxcosx-cos^2x+5 = 0`

`4(1-cos2x)/2+6sin2x-(1+cos2x)/2+5 = 0`

`12sin2x-5cos2x+13 = 0`

`5cos2x-12sin2x = 13`

`(5/13)cos2x-(12/13)sin2x = 1`

 

`5^2+12^2 = 13^2`

If 5,12 and 13 represent respective legs of a right triangle we can say;

`sinalpha = 12/13`

`cosalpha = 5/13 rarr alpha = cos^(-1)(5/13) = 0.374pi`

 

`(5/13)cos2x-(12/13)sin2x = 1`

`cos2xcosalpha-sin2xsinalpha = 1`

`cos(2x+alpha) = cos0`

`2x+alpha = 2npi+-0` where `n in Z`

`x = npi-(alpha)/2`

`x = npi-(0.374pi)/2`

`x = npi-0.187pi`

 

When n = -1 then x <0

when n = 0 then` x <0`

When n = 1 then` x = pi-0.187pi = 0.813pi`

When n = 2 then `x = 2pi-0.187pi = 1.813pi`

When n = 3 then `x > 2pi`

 

So the answers are `x = 0.813pi and x = 1.813pi`

 

 

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