What is the general solution for 1*sin(35)=1.33*sin(θ) ?

The general solution for 1*sin(35)=1.33*sin(θ) is θ=0.44588…+360n,θ=180-0.44588…+360n

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1sin(35)=1.33sin(θ)

1 \cdot sin (3 5^{\circ}) = 1.33 \cdot sin (θ)

θ = 0.44588 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.44588 \dots + 36 0^{\circ} n

Radians

θ = 0.44588 \dots + 2 πn, θ = π - 0.44588 \dots + 2 πn

Solution steps

1 \cdot sin (3 5^{\circ}) = 1.33 sin (θ)

Switch sides

1.33 sin (θ) = 1 \cdot sin (3 5^{\circ})

Multiply:

1 \cdot sin (3 5^{\circ}) = sin (3 5^{\circ})

1.33 sin (θ) = sin (3 5^{\circ})

Divide both sides by

1.33

1.33 sin (θ) = sin (3 5^{\circ})

Divide both sides by

1.33

\frac{1.33 sin ( θ )}{1.33} = \frac{sin ( 3 5 ^{\circ} )}{1.33}

Simplify

sin (θ) = \frac{sin ( 3 5 ^{\circ} )}{1.33}

sin (θ) = \frac{sin ( 3 5 ^{\circ} )}{1.33}

Apply trig inverse properties

sin (θ) = \frac{sin ( 3 5 ^{\circ} )}{1.33}

General solutions for

sin (θ) = \frac{sin ( 3 5 ^{\circ} )}{1.33}

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

θ = arcsin (\frac{sin ( 3 5 ^{\circ} )}{1.33}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{sin ( 3 5 ^{\circ} )}{1.33}) + 36 0^{\circ} n

θ = arcsin (\frac{sin ( 3 5 ^{\circ} )}{1.33}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{sin ( 3 5 ^{\circ} )}{1.33}) + 36 0^{\circ} n

Show solutions in decimal form

θ = 0.44588 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.44588 \dots + 36 0^{\circ} n

cos (3 A) = 1

cos (θ) = (- 0.68159)

cot (x + \frac{5 π}{6}) = 3

θ, y = sin (θ)

sec^{2} (x) - 4 tan^{2} (x) = 0

What is the general solution for 1*sin(35)=1.33*sin(θ) ?
The general solution for 1*sin(35)=1.33*sin(θ) is θ=0.44588…+360n,θ=180-0.44588…+360n