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GarryVolchara [31]
4 years ago
13

Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

Mathematics
1 answer:
Juliette [100K]4 years ago
6 0
First we move sin4x to right side
sin(2x) = sin(4x) 
now we simplify 4x
<span>sin(2x) = sin(2*2x) 
</span>using double angle formula 
<span>sin(2x) = 2sin(2x)cos(2x) 
</span>subtracting sin2x on both sides 
<span>2sin(2x)cos(2x) - sin(2x) = 0 
</span>taking sin2x common
<span>sin(2x) * (2cos(2x) - 1) = 0 
</span>now using product rule 
so 
<span>sin(2x) = 0
or
2cos(2x) - 1 = 0
</span><span>cos(2x) = 1/2</span><span>

hence </span>
<span>x = 0 + nπ/2, π/6 + nπ, 5π/6 + nπ </span>

x = 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6, 2π
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