# Solve for x (4cosx-sinx)^2+5(4cosx-sinx)+4=0

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There are **4** zeros in each period:

Let `a=4cosx-sinx` : then we have `a^2+5a+4=0 ==> a=-4,-1`

(1) `4cosx-sinx=-1` (Note that `cos^2x+sin^2x=1)`

`(4cosx-sinx)^2=(-1)^2`

`16cos^2x-8sinxcosx+sin^2x=1(cos^2x+sin^2x)`

`15cos^2x-8sinxcosx=0`

`cosx(15cosx-8sinx)=0`

So `cosx=0 ==> x=pi/2+npi` . However, squaring introduced extraneous solutions so the only solution that works in the original is `x=pi/2+2npi`

Also `15cosx-8sinx=0`

`tanx=15/8 ==> x=tan^(-1)(15/8)+npi` . Again extraneous solutions were introduced; the only solutions to the original occur at `x=tan^(-1)(15/8)+(2n+1)pi`

(2) `4cosx-sinx=-4`

`(4cosx-sinx)^2=(-4)^2`

`16cos^2x-8sinxcosx+sin^2x=16(cos^2x+sin^2x)`

`15sin^2x+8sinxcosx=0`

`sinx(15sinx+8cosx)=0`

`sinx=0 ==> x=0+npi` but extraneous solutions were introduced so `x=pi+2npi`

`15sinx+8cosx=0==>tanx=-8/15==>x=tan^(-1)(-8/15)` . Again extraneous solutions were introduced so `x=tan^(-1)(-8/15)+(2n+1)pi`

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**The complete solution is:**

`x=pi/2+2npi`

`x=pi+2npi`

`x=tan^(-1)(15/8)+(2n+1)pi` equivalent to `2tan^(-1)(-5/3)+2npi`

`x=tan^(-1)(-8/15)+(2n+1)pi` equivalent to `2tan^(-1)(4)+2npi`

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**Sources:**

You should notice that you may come up with the notation `4cos x- sin x = y ` such that:

`y^2 + 5y + 4 = 0`

You need to solve for y the equation `y^2 + 5y + 4 = 0 ` such that:

`y_(1,2) = (-5+-sqrt(25-16))/2`

`y_(1,2) = (-5+-sqrt9)/2`

`y_(1,2) = (-5+-3)/2`

`y_1 = -1 ; y_2 = -4`

You need to solve the equations `4cos x- sin x = -1` and `4cos x- sin x = -4` such that:

`4cos x- sin x = -1`

You need to use the following formulas:

`sin x = 2sin(tan(x/2))/(1+tan^2(x/2))`

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

You should come up with the notatiun `tan (x/2) = u` such that:

`sin x = (2u)/(1+u^2)`

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

Substsituting `(2u)/(1+u^2)` for sin x and `(1-u^2)/(1+u^2)` for cos x yields:

`4(1-u^2)/(1+u^2) -(2u)/(1+u^2) = -1`

`4(1-u^2) - 2u = -1 - u^2`

`4 - 4u^2 - 2u + 1 + u^2 = 0`

`-3u^2 - 2u + 5 = 0`

`3u^2 + 2u - 5 = 0`

You need to use quadratic formula such that:

`u_(1,2) = (-2+-sqrt(4 + 60))/6 `

`u_(1,2) = (-2+-sqrt(64))/6 `

`u_(1,2) = (-2+-8)/6`

`u_1 = 1 ; u_2 = -5/3`

`tan (x/2) = u =gt tan (x/2) = 1 =gt x/2 = pi/4 + npi`

`x = pi/2 + 2npi`

`tan (x/2) = -5/3 =gt x = 2tan^(-1)(-5/3) + 2npi`

You need to solve for u the equation `4(1-u^2)/(1+u^2) - (2u)/(1+u^2) = -4` `4 - 4u^2 - 2u = -4 - 4u^2`

`8 - 2u = 0 =gt -2u = -8 =gt u = 4`

`tan(x/2) = 4 =gt x = 2tan^(-1)(4) + 2npi`

**Hence, evaluating solutions to trigonometric equations yields `x = pi/2 + 2npi ; x = 2tan^(-1)(-5/3) + 2npi ; x = 2tan^(-1)(4) + 2npi.` **

**Sources:**