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

Radians

θ = 0.56055 \dots + 2 πn, θ = π - 0.56055 \dots + 2 πn

Solution steps

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

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

sin (4 5^{\circ})

Use the following trivial identity:

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

sin (4 5^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \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

1.33 sin (θ) = 1 \cdot \frac{2}{2}

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

1.33 sin (θ) \cdot 100 = 1 \cdot \frac{2}{2} \cdot 100

Refine

133 sin (θ) = 502

133 sin (θ) = 502

Divide both sides by

133

133 sin (θ) = 502

Divide both sides by

133

\frac{133 sin ( θ )}{133} = \frac{50 2}{133}

Simplify

sin (θ) = \frac{50 2}{133}

sin (θ) = \frac{50 2}{133}

Apply trig inverse properties

sin (θ) = \frac{50 2}{133}

General solutions for

sin (θ) = \frac{50 2}{133}

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

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

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