& when sin2e=-2cose
Certainly you mean 0 ≤ x < 360.
cos(x) = 1 => x = 360k, where k is an integer
cos(3x) = 1 => 3x = 0,360 => x = 0,120
sin(2x) = -2cos(x)
2sin(x)cos(x) = -2cos(x)
There is a solution when cos(x) = 0, x = 90,270.
2sin(x) = -2
sin(x) = -1, x = 270
So solutions are 90 and 270.
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Certainly you mean 0 ≤ x < 360.
cos(x) = 1 => x = 360k, where k is an integer
cos(3x) = 1 => 3x = 0,360 => x = 0,120
sin(2x) = -2cos(x)
2sin(x)cos(x) = -2cos(x)
There is a solution when cos(x) = 0, x = 90,270.
2sin(x) = -2
sin(x) = -1, x = 270
So solutions are 90 and 270.