Question

How do you solve the equation on the interval [0,2pi) for `(cos(t))^2=3/4`?

Answer 1

Solution: In interval `[0,2pi) , t= pi/6, t=(5pi)/6, t=(7pi)/6 , t=(11pi)/6`

Explanation:

`(cos(t))^2=3/4 or cos(t)= +-sqrt3/2` we know `cos(pi/6)=sqrt3/2`

When `cos(t)= sqrt3/2 ,t=pi/6 and t=(2pi-pi/6) or t=(11pi)/6` AND

When `cos(t)= -sqrt3/2 ,t=(pi-pi/6)=(5pi)/6 and t=(pi+pi/6) =(7pi)/6`

Solution: In interval `[0,2pi) , t= pi/6, t=(5pi)/6, t=(7pi)/6 , t=(11pi)/6` [Ans]