What is the general solution for -15.44=-2sin(-3θ+300) ?

The general solution for -15.44=-2sin(-3θ+300) is No Solution for θ\in\mathbb{R}

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-15.44=-2sin(-3θ+300)

- 15.44 = - 2 sin (- 3 θ + 300)

No Solution for θ \in R

Solution steps

- 15.44 = - 2 sin (- 3 θ + 300)

Switch sides

- 2 sin (- 3 θ + 300) = - 15.44

Divide both sides by

- 2

- 2 sin (- 3 θ + 300) = - 15.44

Divide both sides by

- 2

\frac{- 2 sin ( - 3 θ + 300 )}{- 2} = \frac{- 15.44}{- 2}

Simplify

\frac{- 2 sin ( - 3 θ + 300 )}{- 2} = \frac{- 15.44}{- 2}

Simplify

\frac{- 2 sin ( - 3 θ + 300 )}{- 2} : sin (- 3 θ + 300)

\frac{- 2 sin ( - 3 θ + 300 )}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 sin ( - 3 θ + 300 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin (- 3 θ + 300)

Simplify

\frac{- 15.44}{- 2} : 7.72

\frac{- 15.44}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{15.44}{2}

Divide the numbers:

\frac{15.44}{2} = 7.72

= 7.72

sin (- 3 θ + 300) = 7.72

sin (- 3 θ + 300) = 7.72

sin (- 3 θ + 300) = 7.72

- 1 \leq sin (x) \leq 1

No Solution for θ \in R

sin (x) = - cos (3 4^{\circ})

2 sin^{2} (x) = cos^{3} (x) tan (x)

tan (x - 1 0^{\circ}) cot (2 0^{\circ} - x) = 1

3 csc (2 x) - 4 sin (2 x) = 0

3 tan (B) - 4 = tan (B) - 2

What is the general solution for -15.44=-2sin(-3θ+300) ?
The general solution for -15.44=-2sin(-3θ+300) is No Solution for θ\in\mathbb{R}