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

##### 1 Answer

Aug 7, 2016

#### 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

The general value is

n =1 and for negative sign,