Posted by **Calvin** on Thursday, May 17, 2012 at 4:47pm.

solve 2cos^2x-sinx=1, 0 < or equal to x < 2pie

- trig -
**Bosnian**, Thursday, May 17, 2012 at 5:33pm
cos ^ 2 ( x ) = 1 - sin ^ 2 ( x )

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

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

2 - 2 sin ^ 2 ( x ) - sin ( x ) = 1 Subtract 1 to both sides

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

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

Substitute :

sin ( x ) = u

- 2 u ^ 2 - u + 1 = 0

The exact solution are :

u = 1 / 2

u = - 1

cos ( x ) = 1 / 2

cos ( x ) = - 1

x = 30 ° = pi / 6

x = 270 ° = 9 pi / 6 rad

P.S

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- 2 u ^ 2 - u + 1 = 0

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- 2 u ^ 2 - u + 1 = 0

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- trig -
**Bosnian**, Thursday, May 17, 2012 at 5:34pm
Corection:

sin ( x ) = 1 / 2

sin ( x ) = - 1

- trig -
**ky**, Thursday, May 17, 2012 at 5:43pm
would the 3 answers be: pi/6, 5pi/6, 3pi/2

- trig -
**Bosnian**, Thursday, May 17, 2012 at 6:00pm
Yes

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