Posted by **Anonymous** on Sunday, November 22, 2009 at 1:11pm.

Determine all solutions in the interval

x is all real numbers , [0, 2 pi]

using a trigonometric identify

2cos^2x + sinx - 1 = 0

- Advanced Functions -
**Reiny**, Sunday, November 22, 2009 at 4:33pm
2cos^2x + sinx - 1 = 0

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

2 - 2sin^2x + sinx - 1 = -

2sin^2x - sinx -1 = 0

(2sinx + 1)(sinx - 1) = 0

sinx = -1/2 or sinx = 1

from sinx = -1/2, x = pi + pi/6 or 2pi - pi/6

x = 7pi/6 radians or 11pi/6 radians ,

(210º or 330º)

from sinx = 1, x = pi/4 radians, (90º)

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