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How solve equation sin2x+5(sinx+cosx)=-1?

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greenbel | (Level 2) Honors

Posted May 27, 2013 at 1:19 PM via web

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How solve equation sin2x+5(sinx+cosx)=-1?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 27, 2013 at 1:46 PM (Answer #1)

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You should replace `sin 2x` by the equivalent product such that:

`2sin x*cos x + 5(sin x + cos x) = -1`

You should come up with the substitution, such that:

`sin x + cos x = y => (sin x + cos x)^2 = y^2 `

Squaring yields:

`sin^2 x + 2sin x*cos x + cos^2 x = y^2`

Replacing 1 for `sin^2 x + cos^2 x` yields:

`1 + 2sin x*cos x = y^2 => 2sin x*cos x = y^2 - 1`

`sin x*cos x = (y^2 - 1)/2`

Changing the variable in equation yields:

`2(y^2 - 1)/2 + 5y = -1`

Reducing duplicate terms yields:

`y^2 - 1 + 5y = -1 => y^2 + 5y = 0`

Factoring out y yields:

`y(y + 5) = 0`

Using zero product rule yields:

`y = 0`

`y + 5 = 0 => y = -5`

Replacing back `sin x + cos x` for y yields:

`sin x + cos x = 0 => sin x/cos x + 1 = 0 => tan x + 1 = 0`

`tan x = -1 => x = -pi/4 + n*pi`

`sin x + cos x= -5` invalid since `|sin x + cos x| <= sqrt 2`

Hence, evaluating the solution to the given equation yields `x = -pi/4 + n*pi.`

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