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…

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

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

t = 1.96558 \dots

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

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

31.91 cos (x) = 20.48

7 sin (x) + 7 sin (x) tan (x) = 0

0 = cos (20 t) - cos (20.02 t)

2 tan (2 x - 10) = 3.4641

12 cos (\frac{π}{3} x) + 8 = 0

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…