What is the general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) ?

The general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) is θ=1.45945…+2pin,θ=2pi-1.45945…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(θ)= 1/(sqrt(1^2+(4)^2+8^2))

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

Radians

θ = 1.45945 \dots + 6.28318 \dots n, θ = 4.82372 \dots + 6.28318 \dots n

Solution steps

cos (θ) = \frac{1}{1 ^{2} + ( 4 ) ^{2} + 8 ^{2}}

Simplify

\frac{1}{1 ^{2} + 4 ^{2} + 8 ^{2}} : \frac{1}{9}

\frac{1}{1 ^{2} + 4 ^{2} + 8 ^{2}}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{1 + 4 ^{2} + 8 ^{2}}

1 + 4^{2} + 8^{2} = 81

1 + 4^{2} + 8^{2}

4^{2} = 16

= 1 + 16 + 8^{2}

8^{2} = 64

= 1 + 16 + 64

Add the numbers:

1 + 16 + 64 = 81

= 81

= \frac{1}{81}

81 = 9

81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

= \frac{1}{9}

Apply trig inverse properties

cos (θ) = \frac{1}{9}

General solutions for

cos (θ) = \frac{1}{9}

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

θ = arccos (\frac{1}{9}) + 2 πn, θ = 2 π - arccos (\frac{1}{9}) + 2 πn

θ = arccos (\frac{1}{9}) + 2 πn, θ = 2 π - arccos (\frac{1}{9}) + 2 πn

Show solutions in decimal form

θ = 1.45945 \dots + 2 πn, θ = 2 π - 1.45945 \dots + 2 πn

2 + 3 cot^{2} (x) - 3 csc (x) = 5

- 2 = - 2 + tan (θ)

cos (A) = 0.65

sin (3 x - \frac{π}{4}) = \frac{2}{2}

sin (2 θ) = \frac{- 3}{4}, \frac{3 π}{2} \leq 2 θ \leq 2 π

What is the general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) ?
The general solution for cos(θ)= 1/(sqrt(1^2+(4)^2+8^2)) is θ=1.45945…+2pin,θ=2pi-1.45945…+2pin