What is the general solution for (179)/(sin(x+20))=(125)/(sin(125)) ?

The general solution for (179)/(sin(x+20))=(125)/(sin(125)) is No Solution for x\in\mathbb{R}

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(179)/(sin(x+20))=(125)/(sin(125))

\frac{179}{sin ( x + 2 0 ^{\circ} )} = \frac{125}{sin ( 12 5 ^{\circ} )}

No Solution for x \in R

Solution steps

\frac{179}{sin ( x + 2 0 ^{\circ} )} = \frac{125}{sin ( 12 5 ^{\circ} )}

Cross multiply

\frac{179}{sin ( x + 2 0 ^{\circ} )} = \frac{125}{sin ( 12 5 ^{\circ} )}

Apply fraction cross multiply: if

\frac{a}{b} = \frac{c}{d}

then

a \cdot d = b \cdot c

179 sin (12 5^{\circ}) = sin (x + 2 0^{\circ}) \cdot 125

179 sin (12 5^{\circ}) = sin (x + 2 0^{\circ}) \cdot 125

Switch sides

sin (x + 2 0^{\circ}) \cdot 125 = 179 sin (12 5^{\circ})

Divide both sides by

125

sin (x + 2 0^{\circ}) \cdot 125 = 179 sin (12 5^{\circ})

Divide both sides by

125

\frac{sin ( x + 2 0 ^{\circ} ) \cdot 125}{125} = \frac{179 sin ( 12 5 ^{\circ} )}{125}

Simplify

sin (x + 2 0^{\circ}) = \frac{179 sin ( 12 5 ^{\circ} )}{125}

sin (x + 2 0^{\circ}) = \frac{179 sin ( 12 5 ^{\circ} )}{125}

Verify Solutions

Find undefined (singularity) points:

sin (x + 2 0^{\circ}) = 0

Take the denominator(s) of

\frac{179}{sin ( x + 2 0 ^{\circ} )}

and compare to zero

sin (x + 2 0^{\circ}) = 0

The following points are undefined

sin (x + 2 0^{\circ}) = 0

Combine undefined points with solutions:

sin (x + 2 0^{\circ}) = \frac{179 sin ( 12 5 ^{\circ} )}{125}

- 1 \leq sin (x) \leq 1

No Solution for x \in R

sin (x) = \frac{15}{12}

sin (z) = 5

cos (5 x) cos (x) + sin (5 x) sin (x) = 0

sin (θ) = \frac{5 n + 1}{3} - 2 n

cosh (x) - 1 = 0

What is the general solution for (179)/(sin(x+20))=(125)/(sin(125)) ?
The general solution for (179)/(sin(x+20))=(125)/(sin(125)) is No Solution for x\in\mathbb{R}