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find all solutions to sin^2x+cosx-1=0

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tarfa123 | Student, Undergraduate | (Level 1) eNoter

Posted March 2, 2012 at 10:51 PM via web

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find all solutions to sin^2x+cosx-1=0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 2, 2012 at 10:59 PM (Answer #1)

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You need to write the equation in terms of cos x only, hence you should use the basic formula of trigonometry `sin^2 x + cos^2 x = 1` .

Substituting the constant term 1 by thesum of squares `sin^2 x + cos^2 x`  yields:

`sin^2 x + cos x - (sin^2 x + cos^2 x)=0`

Opening the brackets yields:

`sin^2 x + cos x - sin^2 x- cos^2 x = 0`

Reducing like terms yields:

`cos x - cos^2 x = 0`

You need to factor out cos x such that:

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

`x = +-cos^(-1) (0) + 2npi`

`x = +-pi/2 + 2npi`

`1 - cos x = 0 =gt -cos x = -1`

`cos x = 1 =gt x = +-cos^(-1) (1) + 2npi`

`x = 2npi`

Hence, evaluating all solutions to equation `sin^2 x + cos x - 1= 0`  yields `x = +-pi/2 + 2npi ` and `x = 2npi.`

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

Posted March 3, 2012 at 12:53 AM (Answer #2)

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Find all solutions to `sin^2x+cosx-1=0` :

Use the pythagorean identity to rewrite `sin^2x` as `1-cos^2x`

Thus `1-cos^2x+cosx-1=0` or

`cos^2x-cosx=0`

`cosx(cosx-1)=0`

`=>cosx=0` or `cosx=1`

If `cosx=0` then `x=pi/2+npi` for `ninZZ` (n an integer)

If `cosx=1` then `x=0+2npi` for `ninZZ`

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tonys538 | TA , Undergraduate | (Level 1) Valedictorian

Posted December 31, 2014 at 5:22 PM (Answer #3)

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The equation sin^2x+cosx-1=0 has to be solved for x.

sin^2x+cosx-1=0

Replace sin^2x with a square of cos x using the relation sin^2x = 1 - cos^2x

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

-cos^2x + cos x = 0

cos x(-cos x + 1) = 0

cos x = 0 is true for all values of x = 90 + n*360 and 270 + n*360

-cos x + 1 = 0

cos x = 1

This is true for all values of x = n*360 and 180 + n*360

The roots of the equation sin^2x+cosx-1=0 are n*360, 90+n*360, 180 +n*260 and 270 + n*360 degrees.

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