What is the general solution for sin(2θ)=-1/8 ,270<θ<360 ?

The general solution for sin(2θ)=-1/8 ,270<θ<360 is θ=(540+0.12532…)/2 ,θ=(-0.12532…+720)/2

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sin(2θ)=-1/8 ,270<θ<360

sin (2 θ) = - \frac{1}{8}, 27 0^{\circ} < θ < 36 0^{\circ}

θ = \frac{54 0 ^{\circ} + 0.12532 \dots}{2}, θ = \frac{- 0.12532 \dots + 72 0 ^{\circ}}{2}

Solution steps

sin (2 θ) = - \frac{1}{8}, 27 0^{\circ} < θ < 36 0^{\circ}

Apply trig inverse properties

2 θ = arcsin (- \frac{1}{8}) + 36 0^{\circ} n, 2 θ = 18 0^{\circ} + arcsin (\frac{1}{8}) + 36 0^{\circ} n

Solve

2 θ = arcsin (- \frac{1}{8}) + 36 0^{\circ} n : θ = - \frac{arcsin ( \frac{1}{8} )}{2} + 18 0^{\circ} n

Solve

2 θ = 18 0^{\circ} + arcsin (\frac{1}{8}) + 36 0^{\circ} n : θ = 9 0^{\circ} + \frac{arcsin ( \frac{1}{8} )}{2} + 18 0^{\circ} n

θ = - \frac{arcsin ( \frac{1}{8} )}{2} + 18 0^{\circ} n, θ = 9 0^{\circ} + \frac{arcsin ( \frac{1}{8} )}{2} + 18 0^{\circ} n

Solutions for the range

27 0^{\circ} < θ < 36 0^{\circ}

θ = \frac{54 0 ^{\circ} + arcsin ( \frac{1}{8} )}{2}, θ = \frac{- arcsin ( \frac{1}{8} ) + 72 0 ^{\circ}}{2}

Show solutions in decimal form

θ = \frac{54 0 ^{\circ} + 0.12532 \dots}{2}, θ = \frac{- 0.12532 \dots + 72 0 ^{\circ}}{2}

1 - 2 cos^{2} (x) - cos (x) = 0

\frac{1}{3} = cos (\frac{π x}{12})

\frac{cos ( 3 0 ^{\circ} )}{cos ( 6 0 ^{\circ} )} = tan (x)

4 cos^{2} (x) - 18 cos (x) - 10 = 0

25 sin (θ) - 1.5 cos (θ) = 20

What is the general solution for sin(2θ)=-1/8 ,270<θ<360 ?
The general solution for sin(2θ)=-1/8 ,270<θ<360 is θ=(540+0.12532…)/2 ,θ=(-0.12532…+720)/2