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Question

Solve sin3x<sinx

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Solution

sin3x<sinx
sin3xsinx<0
Put sin3x=3sinx4sin3x
3sinx4sin3xsinx<0
2sinx4sin3x<0
2sinx(12sin2x)<0
sinxcos2x<0 [cos2x=12sin2x]
sinx+ve cos2xve
xϵ(π4,π2][π2,3π4)=(π4,3π4)
xve cos2x+ve
xϵ(π,5π4)(7π4,2π)
General solution will be xϵ(xπ+π4+nπ+3π4)(nπ+π,nπ+5π4)(nπ+7π4+nπ+2π)
Where n is even number.

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