Help me to solve this math trig
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Help me to solve this math trig

[From: ] [author: ] [Date: 11-05-14] [Hit: ]
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2sinx(1 + cos2x) + sin2x = 1 + cos2x

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2sinx(1 + cos2x) + sin2x = 1 + cos2x
<=> 2sinx(1 + 2cos²x - 1 ) + 2sinxcosx = 1 + 2cos²x - 1
<=> 2sinx(2cos²x) + 2sinxcosx - 2cos²x = 0
<=> 4sinxcos²x + 2sinxcosx - 2cos²x = 0
<=> 2cosx( 2sinxcosx + sinx - cosx ) = 0

<=> cosx = 0 --> x = π/2 + kπ ( k of Z )
<=> 2sinxcosx + sinx - cosx = 0
Change sinx - cosx = t ( |t| ≤ √2 )
<=> (sinx - cosx)² = t²
<=> sin²x - 2sinxcosx + cos²x = t²
<=> 1 - 2sinxcosx = t²
<=> 2sinxcosx = 1 - t²

=> 1 - t² + t = 0
<=> -t² + t + 1 = 0

<=> t = (1 + √5)/2
<=> t = (1 - √5)/2

<=> √2.sin(x - π/4) = (1 + √5)/2
<=> √2.sin(x - π/4) = (1 - √5)/2

<=> sin(x - π/4) = (√10 + √2)/4
<=> sin(x - π/4) = (-√10 + √2)/4

<=> x - π/4 = arcsin[(-√10 + √2)/4] + k2π ( k of Z )
<=> x - π/4 = π - arcsin[(-√10 + √2)/4] + k2π

<=> x = arcsin[(-√10 + √2)/4] + π/4 + k2π
<=> x = 5π/4 - arcsin[(-√10 + √2)/4] + k2π
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