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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(cos(α-30))/1 =(sin(60))/3

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

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}

Solution steps

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

sin (6 0^{\circ}) = \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

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}

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 = 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}

tan (x) tan (7 x) = 1, x

3 sec^{2} (x) + tan (x) - 6 = 0

2 sin (x) + cot (x) = csc (x)

8 cos^{2} (x) + 2 cos (x) - 3 = 0

cos (A) = \frac{1}{65}, cos (B) = \frac{1}{5}

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