Homework Help

What is x if `sqrt(1+sinx)` +`sqrt(1-sinx)` =(`sqrt2` )sinx?

user profile pic

minlux | Honors

Posted September 27, 2013 at 4:16 PM via web

dislike 1 like

What is x if `sqrt(1+sinx)` +`sqrt(1-sinx)` =(`sqrt2` )sinx?

Tagged with math, trigonometry, x

1 Answer | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 27, 2013 at 4:29 PM (Answer #1)

dislike 0 like

You need to square both sides, such that:

`(sqrt(1 + sin x) + sqrt(1 - sin x))^2 = (sqrt2*sin x)^2`

You need to expand the square to the left, such that:

`1 + sin x + 2*sqrt((1 - sin x)(1 + sin x)) + 1 - sin x = 2sin^2 x`

Reducing duplicate members, yields:

`2 + 2*sqrt((1 - sin x)(1 + sin x)) = 2sin^2 x`

You may convert the product `(1 - sin x)(1 + sin x)` into a difference of squares, such that:

`2 + 2*sqrt(1 - sin^2 x) = 2sin^2 x`

Using Pythagorean identity, yields:

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

`2 + 2*sqrt(cos^2 x) = 2sin^2 x`

Factoring out 2 yields:

`2(1 + sqrt(cos^2 x)) = 2sin^2 x`

You need to divide by 2 both sides, such that:

`1 + sqrt(cos^2 x) = sin^2 x`

`1 + |cos x| = sin^2 x`

You need to replace `1 - cos^2 x` for `sin^2 x` , such that:

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

You need to consider `|cos x| = cos x` if `cos x >=0` , such that:

`1 + cos x = 1 - cos^2 x => cos x + cos^2 x = 0`

Factoring out `cos x` , yields:

`cos x(1 + cos x) = 0 `

Using the zero product rule, yields:

`cos x = 0 => x = +-pi/2 + 2npi`

`1 + cos x = 0 => cos x = -1` invalid under the imposed condition, `cos x >= 0` .

You need to consider `|cos x| = cos x` if `cos x<0` , such that:

`1 - cos x = 1 - cos^2 x => cos^2 x - cos x = 0`

`cos x(cos x -1) = 0`

Using zero product rule, yields:

`cos x = 0` , invalid under the imposed condition, `cos x < 0` .

`cos x - 1 = 0 => cos x = 1` , invalid under the imposed condition, `cos x < 0` .

You need to check the values `x = +-pi/2` in equation, such that:

`x = pi/2 => sqrt(1 + sin (pi/2)) + sqrt(1 - sin (pi/2)) = sqrt2*sin(pi/2)`

`sqrt(1 + 1) + sqrt(1 - 1) = sqrt 2 => sqrt 2 + sqrt 0 = sqrt 2 => sqrt2 = sqrt2`

`x = -pi/2 => sqrt(1 + sin (-pi/2)) + sqrt(1 - sin (-pi/2)) = sqrt2*sin(-pi/2)`

`sqrt(1 - 1) + sqrt(1 + 1 ) = -sqrt2 => sqrt2 != -sqrt2`

Hence, evaluating the general solution to the given trigonometric equation, yields `x = pi/2 + 2npi.`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes