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

Question

2265 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-07T09:05:25

$218.94^o$ ,

#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

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

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