Round 124,567 to nearest thousand
1 answer:
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The solutions appear to be {π/2, 2π/3, 4π/3}.
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Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2
and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
_____
Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2
and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}