What is the general solution for (cos(-30+x))/(cos(30+x))= 1835/726 ?

The general solution for (cos(-30+x))/(cos(30+x))= 1835/726 is x=0.64352…+180n

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(cos(-30+x))/(cos(30+x))= 1835/726

\frac{cos ( - 3 0 ^{\circ} + x )}{cos ( 3 0 ^{\circ} + x )} = \frac{1835}{726}

x = 0.64352 \dots + 18 0^{\circ} n

Solution steps

\frac{cos ( - 3 0 ^{\circ} + x )}{cos ( 3 0 ^{\circ} + x )} = \frac{1835}{726}

Rewrite using trig identities

\frac{\frac{3}{2} cos ( x ) + \frac{1}{2} sin ( x )}{\frac{3}{2} cos ( x ) - \frac{1}{2} sin ( x )} = \frac{1835}{726}

Subtract

\frac{1835}{726}

from both sides

\frac{\frac{3}{2} cos ( x ) + \frac{1}{2} sin ( x )}{\frac{3}{2} cos ( x ) - \frac{1}{2} sin ( x )} - \frac{1835}{726} = 0

Simplify

\frac{\frac{3}{2} cos ( x ) + \frac{1}{2} sin ( x )}{\frac{3}{2} cos ( x ) - \frac{1}{2} sin ( x )} - \frac{1835}{726} : \frac{- 1109 3 cos ( x ) + 2561 sin ( x )}{726 ( 3 cos ( x ) - sin ( x ) )}

\frac{- 1109 3 cos ( x ) + 2561 sin ( x )}{726 ( 3 cos ( x ) - sin ( x ) )} = 0

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

- 11093 cos (x) + 2561 sin (x) = 0

Rewrite using trig identities

- 11093 + 2561 tan (x) = 0

Move

11093

to the right side

2561 tan (x) = 11093

Divide both sides by

2561

tan (x) = \frac{1109 3}{2561}

Apply trig inverse properties

x = arctan (\frac{1109 3}{2561}) + 18 0^{\circ} n

Show solutions in decimal form

x = 0.64352 \dots + 18 0^{\circ} n

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

4 sin (2 x + \frac{π}{6}) = 2

8 cos (θ) = 3 - 4 cos (θ)

cos (θ) = \frac{3}{2}, sin (θ)

sin (θ) = 0.788

What is the general solution for (cos(-30+x))/(cos(30+x))= 1835/726 ?
The general solution for (cos(-30+x))/(cos(30+x))= 1835/726 is x=0.64352…+180n