Thread: Conditional Trigonometric Equations

Solve in the interval [0,2], answer as a multiple of pi.
The first on is (sin(t))^2=1/2 and her is what I did.
t= sqrt(1/2)= +-sqrt(2)/2
so, t=pi/4 or t=5pi/4

the next one is the same type but uses cosine.
(cos(t))^2=3/4
t= +-sqrt(3)/2
so, t=pi/6 or -5pi/6

When someone has time could they look this over and let me know if I am going in the right direction with these.

Thank you very much!!
Keith Stevens

Originally Posted by kcsteven

Solve in the interval [0,2], answer as a multiple of pi.
The first on is (sin(t))^2=1/2 and her is what I did.
t= sqrt(1/2)= +-sqrt(2)/2
so, t=pi/4 or t=5pi/4

the next one is the same type but uses cosine.
(cos(t))^2=3/4
t= +-sqrt(3)/2
so, t=pi/6 or -5pi/6

When someone has time could they look this over and let me know if I am going in the right direction with these.
Thank you very much!!
Keith Stevens

Hello,

first I've a question: Is the domain really [0, 2]? Or do you mean [0, 2pi]?

The results of the first problem are OK.

According to your domain you can't use negative results. And: 5/6*pi = 2.61... that means, it is not in your domain. The only possible result is pi/6.

EB