cosx + sinx = cos2x+ sin2x
Answers
Answer:
cosx+sinx=cos2x+sin2x, then
cosx-cos2x=sin2x-sinx
2sin(x+2x)/2sin(2x-x)/2=2cos(2x+x)/2sin(2x-x)/2
sin(3x/2) sin(x/2) - cos(3x/2) sin(x/2) = 0
sin(x/2) [sin(3x/2) - cos(3x/2)] = 0
Either sin(x/2) = 0, or sin(3x/2) - cos(3x/2)=0
So if sin(x/2) = 0, then
sin (x/2) = sin 0
Thus x/2 = nπ +(-1)ⁿ ×0
or x/2 = nπ or x = 2nπ.
And if sin(3x/2) - cos(3x/2) = 0
Dividing both sides by cos(3x/2), we get
tan(3x/2) - 1 = 0
tan(3x/2) = 1
tan(3x/2) = tan(π/4)
3x/2 = nπ + π/4
3x = 2nπ + π/2
x = (2nπ/3) + π/6.
Step-by-step explanation:
cosx+sinx=cos2x+sin2x, then
cosx-cos2x=sin2x-sinx
2sin(x+2x)/2sin(2x-x)/2=2cos(2x+x)/2sin(2x-x)/2
sin(3x/2) sin(x/2) - cos(3x/2) sin(x/2) = 0
sin(x/2) [sin(3x/2) - cos(3x/2)] = 0
Either sin(x/2) = 0, or sin(3x/2) - cos(3x/2)=0
So if sin(x/2) = 0, then
sin (x/2) = sin 0
Thus x/2 = nπ +(-1)ⁿ ×0
or x/2 = nπ or x = 2nπ.
And if sin(3x/2) - cos(3x/2) = 0
Dividing both sides by cos(3x/2), we get
tan(3x/2) - 1 = 0
tan(3x/2) = 1
tan(3x/2) = tan(π/4)
3x/2 = nπ + π/4
3x = 2nπ + π/2
x = (2nπ/3) + π/6.
Answer:
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