Solve:(by step)
sin 3x + sin x = 0
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Solve sin3x<sinx
Medium
Solution
verified
Verified by Toppr
sin3x<sinx
sin3x−sinx<0
Put sin3x=3sinx−4sin
3
x
3sinx−4sin
3
x−sinx<0
2sinx−4sin
3
x<0
2sinx(1−2sin
2
x)<0
sinxcos2x<0 [cos2x=1−2sin
2
x]
sinx→+ve cos2x→−ve
xϵ(
4
π
,
2
π
]∪[
2
π
,
4
3π
)=(
4
π
,
4
3π
)
x→−ve cos2x→+ve
xϵ(π,
4
5π
)∪(
4
7π
,2π)
General solution will be xϵ(xπ+
4
π
+nπ+
4
3π
)∪(nπ+π,nπ+
4
5π
)∪(nπ+
4
7π
+nπ+2π)
Where n is even number.
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