Re: Trigonometric Functions

It is odd that the stated interval for is given in radians, yet the roots are given in degrees. Anyway, we are told:

which means:

From the given equation, we find:

hence:

Re: Trigonometric Functions

Ok, thank you. My only other question then is why, based on the 2 answer outcome to the problem in the first post, is the answer to the similar question of for shown as a three solution answer ?

Following, on from above I get:

. Cos is negative in the 2nd & 3rd quadrants.

or

How do I account for the 160?

Re: Trigonometric Functions

With:

we have:

So, we find:

and so:

Re: Trigonometric Functions