How you solve (square root(1+4cosx))sinx=2sin45?

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embizze | High School Teacher | (Level 2) Educator Emeritus

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Solve `sinxsqrt(1+4cosx)=2sin45^@` :

The right hand side is `sqrt(2)` since `sin45^@=sqrt(2)/2`

`sinxsqrt(1+4cosx)=sqrt(2)` Square both sides (we might introduce extraneous solutions so we must check all possible solutions in the original equation.)

`sin^2(x)(1+4cos(x))=2` Use the Pythagorean substitution:

`(1-cos^2x)(1+4cosx)=2` Expand the product:

`1+4cosx-cos^2x-4cos^3x=2`

`4cos^3x+cos^2x-4cosx+1=0` This is a cubic in cosx:

`4w^3+w^2-4w+1=0`

Using an algebra utility we get `w~~-1.229075,.2991394,.67995599`

`cosx=-1.229075` is impossible

`cosx=.2991394==>x~~1.267"rad"`

`cosx=.67995599==>x~~.8231"rad"`

Substituting these values into the original equation gives the correct result.

The answers are `x~~1.267+2npi"or".8231+2npi`

(In degrees `x~~72.59^@,47.16^@` )

The graph:

`y=sqrt(2)` in red.

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