sin2x+sin4x=cosx+cos3x
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Given : sin2x+sin4x=cosx+cos3x
To Find : Values of x in [0 , π/2]
Solution:
sin2x+sin4x=cosx+cos3x
sin2x+sin4x = 2Sin3xCosx
cosx+cos3x = 2Cos2xCosx
=> 2Sin3xCosx = 2Cos2xCosx
=> Sin3xCosx = Cos2xCosx
=> Sin3xCosx - Cos2xCosx = 0
=> cosx (Sin3x - Cos2x) = 0
=> cosx = 0 => x = π/2
or Sin3x = Cos2x
Sinx = Cos(π/2 - x)
3x = π/2 - 2x
=> 5x = π/2
=> x = π/10
π/2 and π/10
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