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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.
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).)
√ from which we get cos(x) = 0 or sin(x) =
√ integers k . From sin(x) =
2, we get x = As always, experience is the greatest teacher here!
As always, when in doubt, write it out! π
2. From cos(x) = 0, we obtain x = +2πk or x = 2π
2 + πk for +2πk for integers k . The answers 736 Foundations of Trigonometry
which lie in [0, 2π ) are x = π , 32 , π and 23 . We graph y = sin(2x) and y =
after some careful zooming, verify...
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