Question

How do you solve `cos(t)=-7/9` with `pi<t<3pi/2`?

Answer 1

`218.94^o`,

Explanation:

#cos t = -7/9 < 0. So, the principal value of t is in Q2. Q3 is (pi. 3/2pi)

The principal value of t is `141.06^o in Q2`.

The general value is

`2npi+- 141.06^o, n = 0, +- 1, +- 2, +-3, ..`, ...

n =1 and for negative sign, `t in Q3` is obtained as

`(360-141.6)^o=218.94^o`,