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`sin^2(x/2)+cosx=1/2` ``Solve this equation for exact values of "x" where...
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High School Teacher
`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`
Posted by hala718 on December 2, 2011 at 10:52 AM (Answer #1)
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.`
Posted by sciencesolve on December 3, 2011 at 2:20 AM (Answer #2)
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