Asked by Anonymous
Determine all solutions in the interval
x is all real numbers , [0, 2 pi]
using a trigonometric identify
2cos^2x + sinx - 1 = 0
x is all real numbers , [0, 2 pi]
using a trigonometric identify
2cos^2x + sinx - 1 = 0
Answers
Answered by
Reiny
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)
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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