What is the general solution for -8sin(2θ+45)=-4,0<= x<360 ?

The general solution for -8sin(2θ+45)=-4,0<= x<360 is θ=52.5,θ=172.5,θ=232.5,θ=352.5

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-8sin(2θ+45)=-4,0<= x<360

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}

θ = 52. 5^{\circ}, θ = 172. 5^{\circ}, θ = 232. 5^{\circ}, θ = 352. 5^{\circ}

Solution steps

- 8 sin (2 θ + 4 5^{\circ}) = - 4, 0 \leq x < 36 0^{\circ}

Divide both sides by

- 8

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

General solutions for

sin (2 θ + 4 5^{\circ}) = \frac{1}{2}

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n, 2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n

Solve

2 θ + 4 5^{\circ} = 3 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n - 7. 5^{\circ}

Solve

2 θ + 4 5^{\circ} = 15 0^{\circ} + 36 0^{\circ} n : θ = 18 0^{\circ} n + 52. 5^{\circ}

θ = 18 0^{\circ} n - 7. 5^{\circ}, θ = 18 0^{\circ} n + 52. 5^{\circ}

Solutions for the range

0 \leq x < 36 0^{\circ}

θ = 52. 5^{\circ}, θ = 172. 5^{\circ}, θ = 232. 5^{\circ}, θ = 352. 5^{\circ}

sin (b) = \frac{10.21 ( sin ( 61.3 6 ^{\circ} ) )}{11.47}

cos (θ) = 0.513

cot (x) = - \frac{24}{7}

sin (θ) = - \frac{2}{3}, 18 0^{\circ} < θ < 27 0^{\circ}

x, R f = e^{c o s (2 x)}

What is the general solution for -8sin(2θ+45)=-4,0<= x<360 ?
The general solution for -8sin(2θ+45)=-4,0<= x<360 is θ=52.5,θ=172.5,θ=232.5,θ=352.5