What is the general solution for cos(α)=sqrt((1+1/5)/2) ?

The general solution for cos(α)=sqrt((1+1/5)/2) is α=0.68471…+2pin,α=2pi-0.68471…+2pin

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cos(α)=sqrt((1+1/5)/2)

cos (α) = \frac{1 + \frac{1}{5}}{2}

α = 0.68471 \dots + 2 πn, α = 2 π - 0.68471 \dots + 2 πn

Degrees

α = 39.23152 \dots^{\circ} + 36 0^{\circ} n, α = 320.76847 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (α) = \frac{1 + \frac{1}{5}}{2}

\frac{1 + \frac{1}{5}}{2} = \frac{3}{5}

\frac{1 + \frac{1}{5}}{2}

Join

1 + \frac{1}{5} : \frac{6}{5}

1 + \frac{1}{5}

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} + \frac{1}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 5 + 1}{5}

1 \cdot 5 + 1 = 6

1 \cdot 5 + 1

Multiply the numbers:

1 \cdot 5 = 5

= 5 + 1

Add the numbers:

5 + 1 = 6

= 6

= \frac{6}{5}

= \frac{\frac{6}{5}}{2}

Apply the fraction rule:

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

= \frac{6}{5 \cdot 2}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{6}{10}

Cancel the common factor:

2

= \frac{3}{5}

cos (α) = \frac{3}{5}

Apply trig inverse properties

cos (α) = \frac{3}{5}

General solutions for

cos (α) = \frac{3}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

α = arccos (\frac{3}{5}) + 2 πn, α = 2 π - arccos (\frac{3}{5}) + 2 πn

α = arccos (\frac{3}{5}) + 2 πn, α = 2 π - arccos (\frac{3}{5}) + 2 πn

Show solutions in decimal form

α = 0.68471 \dots + 2 πn, α = 2 π - 0.68471 \dots + 2 πn

(tan (x)) (sin^{2} (x) - a) (sec (x) - a) = 0, 0 < a < 1

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

sin (\cdot + \frac{π}{2}) = cos (x)

sin (θ) = (0.000069) 2 (500 \cdot 1 0^{- 9})

(2 sin^{2} (x) - 1) (tan (x) - 1) = 0

What is the general solution for cos(α)=sqrt((1+1/5)/2) ?
The general solution for cos(α)=sqrt((1+1/5)/2) is α=0.68471…+2pin,α=2pi-0.68471…+2pin