What is the general solution for (1.33)sin(25)=(1.5)sin(θ) ?

The general solution for (1.33)sin(25)=(1.5)sin(θ) is θ=0.38409…+360n,θ=180-0.38409…+360n

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(1.33)sin(25)=(1.5)sin(θ)

(1.33) sin (2 5^{\circ}) = (1.5) sin (θ)

θ = 0.38409 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.38409 \dots + 36 0^{\circ} n

Radians

θ = 0.38409 \dots + 2 πn, θ = π - 0.38409 \dots + 2 πn

Solution steps

(1.33) sin (2 5^{\circ}) = (1.5) sin (θ)

Switch sides

1.5 sin (θ) = 1.33 sin (2 5^{\circ})

Divide both sides by

1.5

1.5 sin (θ) = 1.33 sin (2 5^{\circ})

Divide both sides by

1.5

\frac{1.5 sin ( θ )}{1.5} = \frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}

Simplify

sin (θ) = \frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}

sin (θ) = \frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}

Apply trig inverse properties

sin (θ) = \frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}

General solutions for

sin (θ) = \frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

θ = arcsin (\frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}) + 36 0^{\circ} n

θ = arcsin (\frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{1.33 sin ( 2 5 ^{\circ} )}{1.5}) + 36 0^{\circ} n

Show solutions in decimal form

θ = 0.38409 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.38409 \dots + 36 0^{\circ} n

cos (y) = - \frac{1}{2}

\frac{2}{1 + sin ( x )} = 3 - 2 sin (x)

x, 4 cos^{2} (x) = 5 + 4 sin (x)

t, y = cos (2 t)

cos (3 x) = \frac{1}{5}

What is the general solution for (1.33)sin(25)=(1.5)sin(θ) ?
The general solution for (1.33)sin(25)=(1.5)sin(θ) is θ=0.38409…+360n,θ=180-0.38409…+360n