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`sin^2(x/2)+cosx=1/2` ``Solve this equation for exact values of "x" where...

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cthirkill | Student, Undergraduate | eNotes Newbie

Posted December 2, 2011 at 10:11 AM via web

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

``

Solve this equation for exact values of "x" where 0<x<360

 

answer should be in degrees

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted December 2, 2011 at 10:52 AM (Answer #1)

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

We will use trigonometric identities to solve.

We know that `cos2x = 1-2sin^2 x`

`==gt cosx = 1-2sin^2 (x/2) `

`==gt 2sin^2 (x/2) = 1-cosx `

`==gt sin^2 (x/2)= (1-cosx)/2 `

`==gt (1-cosx)/2 + cosx = 1/2 `

`==gt 1-cosx + 2cosx = 1 `

`==gt cosx = 0 `

`==gt x = pi/2, 3pi/2`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 3, 2011 at 2:20 AM (Answer #2)

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Use double angle identity to write cos x as `cos 2*(x/2).`

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

Write the equation:

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

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

Subtract 1 both sides:

`-sin^2 (x/2) = 1/2 - 1 =gt -sin^2 (x/2) = -1/2`

`` `sin (x/2) = +- (sqrt2)/2`

`` `x/2 = n*pi/4`  => `x = n*(pi/2)`

`` The angles that check the equation are `x = pi/2; x = (3pi)/2.`

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