What is the general solution for cos(t)=(-5)/(13), 1/2 pi<t<pi ?

The general solution for cos(t)=(-5)/(13), 1/2 pi<t<pi is t=1.96558…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

cos(t)=(-5)/(13), 1/2 pi<t<pi

cos (t) = \frac{- 5}{13}, \frac{1}{2} π < t < π

t = 1.96558 \dots

Degrees

t = 112.61986 \dots^{\circ}

Solution steps

cos (t) = \frac{- 5}{13}, \frac{1}{2} π < t < π

Apply the fraction rule:

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

cos (t) = - \frac{5}{13}

Apply trig inverse properties

cos (t) = - \frac{5}{13}

General solutions for

cos (t) = - \frac{5}{13}

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

t = arccos (- \frac{5}{13}) + 2 πn, t = - arccos (- \frac{5}{13}) + 2 πn

t = arccos (- \frac{5}{13}) + 2 πn, t = - arccos (- \frac{5}{13}) + 2 πn

Solutions for the range

\frac{1}{2} π < t < π

t = arccos (- \frac{5}{13})

Show solutions in decimal form

t = 1.96558 \dots

What is the general solution for cos(t)=(-5)/(13), 1/2 pi<t<pi ?
The general solution for cos(t)=(-5)/(13), 1/2 pi<t<pi is t=1.96558…