solve for 0 < x < 2pi using trigonometric identities
3sinx = 1 +cos2x
3sinx = 1 +cos2x
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3sinx = 1 + cos2x
from identities:
1 = sin²x + cos²x ; cos²x = 1 - sin²x
cos2x = cos²x - sin²x
3sinx = 1 + cos2x
3sinx = (cos²x + sin²x) + (cos²x - sin²x)
3sinx = 2cos²x
3sinx = 2(1-sin²x)
3sinx = 2 - 2sin²x
2sin²x + 3sinx - 2 = 0
(2sinx - 1)(sinx + 2) = 0
first solution:
2sinx - 1 = 0
2sinx = 1
sinx = ½
x = arcsin ½
x = 30° or π/6
or the supplement 210° or 7π/8
2nd solution:
sinx + 2 = 0
sinx = -2 (out of range) invalid
arcsin = -2
from identities:
1 = sin²x + cos²x ; cos²x = 1 - sin²x
cos2x = cos²x - sin²x
3sinx = 1 + cos2x
3sinx = (cos²x + sin²x) + (cos²x - sin²x)
3sinx = 2cos²x
3sinx = 2(1-sin²x)
3sinx = 2 - 2sin²x
2sin²x + 3sinx - 2 = 0
(2sinx - 1)(sinx + 2) = 0
first solution:
2sinx - 1 = 0
2sinx = 1
sinx = ½
x = arcsin ½
x = 30° or π/6
or the supplement 210° or 7π/8
2nd solution:
sinx + 2 = 0
sinx = -2 (out of range) invalid
arcsin = -2
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3sinx=cos^2x+sin^2x+cos^2x-sin^2x
3sinx=2cos^2x
2sin^2x+3sinx-2=0
sinx=(-3-5)/4=-2 x is out the interval (0, 2pi)
sinx=-3+5/4=1/2 x=pi/6, 7pi/8
3sinx=2cos^2x
2sin^2x+3sinx-2=0
sinx=(-3-5)/4=-2 x is out the interval (0, 2pi)
sinx=-3+5/4=1/2 x=pi/6, 7pi/8
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3sinx=1+cos^2x-sin^2x
=1+1-sin^2x-sin^2x
=2-2sin^2x
=>2sin^2x+3sinx-2=0
=1+1-sin^2x-sin^2x
=2-2sin^2x
=>2sin^2x+3sinx-2=0