What is the general solution for 1sin(45)=1.33sin(θ) ?

The general solution for 1sin(45)=1.33sin(θ) is θ=0.56055…+360n,θ=180-0.56055…+360n

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1sin(45)=1.33sin(θ)

1 sin (4 5^{\circ}) = 1.33 sin (θ)

θ = 0.56055 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.56055 \dots + 36 0^{\circ} n

Solution steps

1 \cdot sin (4 5^{\circ}) = 1.33 sin (θ)

sin (4 5^{\circ}) = \frac{2}{2}

1 \cdot \frac{2}{2} = 1.33 sin (θ)

Switch sides

1.33 sin (θ) = 1 \cdot \frac{2}{2}

Multiply both sides by

100

133 sin (θ) = 502

Divide both sides by

133

sin (θ) = \frac{50 2}{133}

Apply trig inverse properties

θ = arcsin (\frac{50 2}{133}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{50 2}{133}) + 36 0^{\circ} n

Show solutions in decimal form

θ = 0.56055 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.56055 \dots + 36 0^{\circ} n

3 cot^{2} (x) = 9

1 2^{(2)} = 1 3^{(2)} + 5^{(2)} - 2 (13) (5) cos (B)

tan (6 t) = 2

sin (2 x) - sin (x) + cos (2 x) + cos (x) = 0

(cos (x) = 1) = (sec (x) = 1)

What is the general solution for 1sin(45)=1.33sin(θ) ?
The general solution for 1sin(45)=1.33sin(θ) is θ=0.56055…+360n,θ=180-0.56055…+360n