# Calculate 2^1/2*sinx+3^1/2*cos2x, tanx=6^1/2/3 .

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Given `tanx=sqrt(6)/3` find the value of `sqrt(2)sinx+sqrt(3)cos2x` :

(1)Draw a right triangle with acute angle x. The side adjacent to x has length 3 and the side opposite has length `sqrt(6)` since `tanx=("sideopposite")/("sideadjacent")`

Then using the pythagorean theorem we get the hypotenuse as `sqrt(15).`

` ` Then `sinx=sqrt(6)/sqrt(15)=sqrt(10)/5` and `cosx=3/sqrt(15)=sqrt(15)/5` .

(2) We use the identity `cos2x=cos^2x-sin^2x`

(3) Then `sqrt(2)sinx+sqrt(3)cos2x=sqrt(2)sqrt(10)/5+sqrt(3)[15/25-10/25]`

`=sqrt(20)/5+sqrt(3)(1/5)`

`=(2sqrt(5))/5+sqrt(3)/5`

`=(2sqrt(5)+sqrt(3))/5` **which is the solution.**

You need to evaluate `sqrt2*sin 2x + sqrt3*cos 2x ` considering `tan x = sqrt6/3` .

You need to use the formula:

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

Hence, you should raise to square `tan x` such that:

`tan^2 x = 6/9 =gt tan^2 x = 2/3`

You need to add 1 both sides such that:

`1 + tan^2 x = 2/3 + 1 =gt 1 + tan^2 x = 5/3`

You need to equate `5/3` and `1/(cos^2 x)` such that:

`5/3=1/(cos^2 x) =gt cos^2 x = 3/5 =gt cos x = +-sqrt(3/5)`

You need to use the basic trigonometric formula to find `sin x` such that:

`sin x = +-sqrt(1 - 3/5) =gt sin x = +-sqrt(2/5)`

You need to remember the formula for `sin 2x` such that:

`sin 2x = 2 sin x*cos x`

You need to substitute `sqrt(2/5)` for `sin x ` and `sqrt(3/5)` for `cos x ` such that:

`sin 2x = 2*sqrt6/5`

You need to remember the formula for `cos 2x ` such that:

`cos 2x = 2cos^2 x - 1`

`cos 2x = 6/5 - 1 =gt cos 2x = 1/5`

You need to substitute `2*sqrt6/5` for `sin 2x` and `1/5` for `cos 2x ` such that:

`2sqrt12/5 + sqrt3/5 = (4sqrt3 + sqrt3)/5`

`2sqrt12/5 + sqrt3/5 = 5sqrt3/5`

`2sqrt12/5 + sqrt3/5 = sqrt3`

**Hence, evaluating `sqrt2*sin 2x + sqrt3*cos 2x ` under given conditions yields `sqrt 3.` **