What is the general solution for 50/33 =(sin(x))/(sin(120-x)) ?

The general solution for 50/33 =(sin(x))/(sin(120-x)) is x=1.38810…+180n

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50/33 =(sin(x))/(sin(120-x))

\frac{50}{33} = \frac{sin ( x )}{sin ( 12 0 ^{\circ} - x )}

x = 1.38810 \dots + 18 0^{\circ} n

Solution steps

\frac{50}{33} = \frac{sin ( x )}{sin ( 12 0 ^{\circ} - x )}

Switch sides

\frac{sin ( x )}{sin ( 12 0 ^{\circ} - x )} = \frac{50}{33}

Rewrite using trig identities

\frac{sin ( x )}{\frac{3}{2} cos ( x ) + \frac{1}{2} sin ( x )} = \frac{50}{33}

Subtract

\frac{50}{33}

from both sides

\frac{sin ( x )}{\frac{3}{2} cos ( x ) + \frac{1}{2} sin ( x )} - \frac{50}{33} = 0

Simplify

\frac{sin ( x )}{\frac{3}{2} cos ( x ) + \frac{1}{2} sin ( x )} - \frac{50}{33} : \frac{16 sin ( x ) - 50 3 cos ( x )}{33 ( 3 cos ( x ) + sin ( x ) )}

\frac{16 sin ( x ) - 50 3 cos ( x )}{33 ( 3 cos ( x ) + sin ( x ) )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

16 sin (x) - 503 cos (x) = 0

Rewrite using trig identities

16 tan (x) - 503 = 0

Move

503

to the right side

16 tan (x) = 503

Divide both sides by

16

tan (x) = \frac{25 3}{8}

Apply trig inverse properties

x = arctan (\frac{25 3}{8}) + 18 0^{\circ} n

Show solutions in decimal form

x = 1.38810 \dots + 18 0^{\circ} n

cot (x) csc (x) = cos (x)

2 tan (x) = 2

0.5 = cos (\frac{π}{6} x)

cos^{2} (x) = \frac{π}{4}

cot^{2} (θ) + csc (θ) = 1, 0 \leq θ < 2 π

What is the general solution for 50/33 =(sin(x))/(sin(120-x)) ?
The general solution for 50/33 =(sin(x))/(sin(120-x)) is x=1.38810…+180n