find the general solution of equation sinx-3sin2x+sin3x=cos x-3 cos 2X+cos3x
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Answer:
Step-by-step explanation:sinx−3sin2x+sin3x=cosx−3cos2x
+cos3x
⇒(sinx+sin3x)−3sin2x−(cosx+cos3x)
+3cos2x=0
⇒2sin(
x+3x
2
)cos(
x−3x
2
)−3sin2x
−2cos(
x+3x
2
)cos(
x−3x
2
)+3cos2x=0
⇒2sin2xcosx−3sin2x−2cos2xcosx
+3cos2x=0
⇒sin2x(2cosx−3)−cos2x(2cosx−3)=0
⇒(2cosx−3)(sin2x−cos2x)=0
⇒cosx=
3
2
, or, sin2x=cos2x
As, cos x
∈
[-1,1]. Thus,
cos x≠
3
2
So,
sin2x=cos2x
⇒tan2x=1
⇒tan2x=tan
π
4
⇒2x=nπ+
π
4
x=
nπ
2
+
π
8
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