{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

# From theorem 103 we know that since the point b x y

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: of y = cos(3x) and y = cos(5x) below bears this out. √ 6. In examining the equation sin(2x) = 3 cos(x), not only do we have diﬀerent circular functions involved, namely sine and cosine, we also have diﬀerent arguments to contend with, namely 2x and x. Using the identity sin(2x) = 2 sin(x) cos(x) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2x) 2 sin(x) cos(x) √ 2 sin(x) cos(x) − 3 cos(x) √ cos(x)(2 sin(x) − 3) √ = √3 cos(x) = 3 cos(x) (Since sin(2x) = 2 sin(x) cos(x).) =0 =0 √ from which we get cos(x) = 0 or sin(x) = √ integers k . From sin(x) = 6 7 3 2, we get x = As always, experience is the greatest teacher here! As always, when in doubt, write it out! π 3 3 2. From cos(x) = 0, we obtain x = +2πk or x = 2π 3 π 2 + πk for +2πk for integers k . The answers 736 Foundations of Trigonometry π π which lie in [0, 2π ) are x = π , 32 , π and 23 . We graph y = sin(2x) and y = 2 3 after some careful zooming, verify...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online