- #1

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## Homework Statement

Solve the equation:

cos2x - sinx = (1/2) for [0,2pi).

## Homework Equations

sin^2x + cos^2 = 1

## The Attempt at a Solution

cos^2(2x) - sin^2x = 1

1 = 1 Infinite Solutions?

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- #1

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Solve the equation:

cos2x - sinx = (1/2) for [0,2pi).

sin^2x + cos^2 = 1

cos^2(2x) - sin^2x = 1

1 = 1 Infinite Solutions?

- #2

CompuChip

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and

[tex]\cos^2(2x) - \sin^2(x) \neq \cos^2(x) + \sin^2(x)[/tex]

Maybe

[tex]\cos(2x) = \cos^2(x) - \sin^2(x)[/tex]

will help?

- #3

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- #4

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How does that have misplaced exponents and variables? Would Writing parentheses help? Im really sorry for the confusion and my mistake.

Cos(2x) - sin(x) = (1/2) for [0,2pi).

Cosine of 2x minus sin of x equals 1 half for the range 0 through 2pie

- #5

- 1,752

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slap yourself.

Ok, rather than the identity your using ... use:

[tex]\cos{2x}=1-2\sin^2 x[/tex]

Just factor and it's solved!!!

Last edited:

- #6

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1 - 2sin^2(x) - sin(x) = (1/2) [-(1/2)]

-2sin^2(x) - sin(x) + (1/2) = 0

Quadratic Formula?

I tried it and got sinx = (1 +/- Sqrroot(5)) / -4)

x = -54 and 18.

Somehow that doesn't seem right because radians aren't that big on problems. Any help?

- #7

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-54 and 18 are the answers in degrees, and they are correct

- #8

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brendanH the question says between 0 and pi. therefore even if you are right -54 doesnt work

- #9

HallsofIvy

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