What is the general solution for cos(θ)=(2*10^{-5})/(54*10^{-6)} ?

The general solution for cos(θ)=(2*10^{-5})/(54*10^{-6)} is θ=1.19138…+2pin,θ=2pi-1.19138…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(θ)=(210^{-5})/(5410^{-6)}

cos (θ) = \frac{2 \cdot 1 0 ^{- 5}}{54 \cdot 1 0 ^{- 6}}

θ = 1.19138 \dots + 2 πn, θ = 2 π - 1.19138 \dots + 2 πn

Radians

θ = 1.19138 \dots + 6.28318 \dots n, θ = 5.09179 \dots + 6.28318 \dots n

Solution steps

cos (θ) = \frac{2 \cdot 1 0 ^{- 5}}{54 \cdot 1 0 ^{- 6}}

Simplify

\frac{2 \cdot 1 0 ^{- 5}}{54 \cdot 1 0 ^{- 6}} : \frac{10}{27}

\frac{2 \cdot 1 0 ^{- 5}}{54 \cdot 1 0 ^{- 6}}

Cancel the common factor:

2

= \frac{1 0 ^{- 5}}{1 0 ^{- 6} \cdot 27}

Apply exponent rule:

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

\frac{1 0 ^{- 5}}{1 0 ^{- 6}} = 1 0^{- 5 - (- 6)}

= \frac{1 0 ^{- 5 - (- 6)}}{27}

Subtract the numbers:

- 5 - (- 6) = 1

= \frac{10}{27}

Apply trig inverse properties

cos (θ) = \frac{10}{27}

General solutions for

cos (θ) = \frac{10}{27}

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

θ = arccos (\frac{10}{27}) + 2 πn, θ = 2 π - arccos (\frac{10}{27}) + 2 πn

θ = arccos (\frac{10}{27}) + 2 πn, θ = 2 π - arccos (\frac{10}{27}) + 2 πn

Show solutions in decimal form

θ = 1.19138 \dots + 2 πn, θ = 2 π - 1.19138 \dots + 2 πn

sin (a) = \frac{9}{41}

sin (x) = \frac{10}{300}

cos^{2} (x) + 2 sin (x) = 2

cos (21 6^{\circ}) = cos^{2} (a) - sin^{2} (a)

csc (- x) - 1 = cot^{2} (x)

What is the general solution for cos(θ)=(2*10^{-5})/(54*10^{-6)} ?
The general solution for cos(θ)=(2*10^{-5})/(54*10^{-6)} is θ=1.19138…+2pin,θ=2pi-1.19138…+2pin