2sin(theta)= 1 - 2cos(theta)?

Solve for theta, restricting the domain to 0 to 360.

Ed I2011-05-27T20:25:42Z

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2 sin T = 1 - 2 cos T
4 sin^2 T = 1 - 4 cos T + 4 cos^2 T
4 - 4 cos^2 T = 1 - 4 cos T + 4 cos^2 T
0 = 8 cos^2 T - 4 cos T - 3
cos T ≈ 0.9114 or cos T ≈ -0.4114

T ≈ 24° 18', 335° 42', or T ≈ 114° 18', 245° 42'

?2016-12-05T06:04:57Z

Convert the oblong equation x² + y² - 2y = 0 to polar type. keep in mind the conversions between Cartesian and polar. r² = x² + y² y = rsin? Now enable's convert. x² + y² - 2y = 0 r² - 2rsin? = 0 r² = 2rsin? Divide by by potential of r. r = 2sin?

Moise Gunen2011-05-27T20:38:59Z

2sin θ = 1-2cos θ
sin θ + cos θ = 1/2
sin θ + sin(π/2- θ) = 1/2
2sin (π/4)cos(θ - π/4) = 1/2
2(√2)/2cos(θ - π/4) = 1/2
cos(θ - π/4) = 1/(2√2)
cos(θ - π/4) = (√2)/4
θ - π/4 = Arccos((√2)/4 ) or θ - π/4 = 2π - Arccos((√2)/4 )
Answer
θ = π/4 + Arccos((√2)/4 )
θ = 9π/4 - Arccos((√2)/4 )

MechEng20302011-05-27T20:31:22Z

2*sin(t) = 1 - 2*cos(t)

sin(t) + cos(t) = 1/2

Squaring both sides:

1 + sin(2t) = 1/4

sin(2t) = -3/4

t = arcsin(-3/4)* (1/2)