What is the general solution for (cos(30))/(cos(60))=tan(x) ?

The general solution for (cos(30))/(cos(60))=tan(x) is x=60+180n

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(cos(30))/(cos(60))=tan(x)

\frac{cos ( 3 0 ^{\circ} )}{cos ( 6 0 ^{\circ} )} = tan (x)

x = 6 0^{\circ} + 18 0^{\circ} n

Radians

x = \frac{π}{3} + πn

Solution steps

\frac{cos ( 3 0 ^{\circ} )}{cos ( 6 0 ^{\circ} )} = tan (x)

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 0^{\circ})

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 0^{\circ})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{1}{2}

= \frac{1}{2}

\frac{\frac{3}{2}}{\frac{1}{2}} = tan (x)

Switch sides

tan (x) = \frac{\frac{3}{2}}{\frac{1}{2}}

Simplify

\frac{\frac{3}{2}}{\frac{1}{2}} : 3

\frac{\frac{3}{2}}{\frac{1}{2}}

Divide fractions:

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

= \frac{3 \cdot 2}{2 \cdot 1}

Refine

= \frac{3 \cdot 2}{2}

Cancel the common factor:

2

= 3

General solutions for

tan (x) = 3

tan (x)

periodicity table with

18 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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 6 0^{\circ} + 18 0^{\circ} n

x = 6 0^{\circ} + 18 0^{\circ} n

What is the general solution for (cos(30))/(cos(60))=tan(x) ?
The general solution for (cos(30))/(cos(60))=tan(x) is x=60+180n