What is the general solution for (cos(α-30))/1 =(sin(60))/3 ?

The general solution for (cos(α-30))/1 =(sin(60))/3 is α=1.27795…+360n+30,α=360-1.27795…+360n+30

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(cos(α-30))/1 =(sin(60))/3

Solution

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{sin ( 6 0 ^{\circ} )}{3}

Solution

α = 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}

Radians

α = 1.27795 \dots + \frac{π}{6} + 2 πn, α = 2 π - 1.27795 \dots + \frac{π}{6} + 2 πn

Solution steps

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{sin ( 6 0 ^{\circ} )}{3}

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

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

= \frac{3}{2}

= \frac{3}{2}

\frac{cos ( α - 3 0 ^{\circ} )}{1} = \frac{\frac{3}{2}}{3}

Apply rule

\frac{a}{1} = a

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

Apply the fraction rule:

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

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

Multiply the numbers:

2 \cdot 3 = 6

cos (α - 3 0^{\circ}) = \frac{3}{6}

Apply trig inverse properties

cos (α - 3 0^{\circ}) = \frac{3}{6}

General solutions for

cos (α - 3 0^{\circ}) = \frac{3}{6}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n, α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n, α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

Solve

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n : α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α - 3 0^{\circ} = arccos (\frac{3}{6}) + 36 0^{\circ} n

Simplify

arccos (\frac{3}{6}) + 36 0^{\circ} n : arccos (\frac{1}{2 3}) + 36 0^{\circ} n

arccos (\frac{3}{6}) + 36 0^{\circ} n

\frac{3}{6} = \frac{1}{2 3}

\frac{3}{6}

Factor

6 : 2 \cdot 3

Factor

6 = 2 \cdot 3

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

Cancel

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

\frac{3}{2 \cdot 3}

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{2 \cdot 3 ^{- \frac{1}{2} + 1}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

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

Apply radical rule:

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

= \frac{1}{2 3}

= \frac{1}{2 3}

= arccos (\frac{1}{2 3}) + 36 0^{\circ} n

α - 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n

Move

3 0^{\circ}

to the right side

α - 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n

Add

3 0^{\circ}

to both sides

α - 3 0^{\circ} + 3 0^{\circ} = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

Simplify

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

Solve

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n : α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

Simplify

36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n : 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

36 0^{\circ} - arccos (\frac{3}{6}) + 36 0^{\circ} n

\frac{3}{6} = \frac{1}{2 3}

\frac{3}{6}

Factor

6 : 2 \cdot 3

Factor

6 = 2 \cdot 3

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

Cancel

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

\frac{3}{2 \cdot 3}

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{1}{2 \cdot 3 ^{- \frac{1}{2} + 1}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

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

Apply radical rule:

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

= \frac{1}{2 3}

= \frac{1}{2 3}

= 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

Move

3 0^{\circ}

to the right side

α - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n

Add

3 0^{\circ}

to both sides

α - 3 0^{\circ} + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

Simplify

α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

α = arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - arccos (\frac{1}{2 3}) + 36 0^{\circ} n + 3 0^{\circ}

Show solutions in decimal form

α = 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}, α = 36 0^{\circ} - 1.27795 \dots + 36 0^{\circ} n + 3 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for (cos(α-30))/1 =(sin(60))/3 ?
The general solution for (cos(α-30))/1 =(sin(60))/3 is α=1.27795…+360n+30,α=360-1.27795…+360n+30