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Unformatted text preview: 2)2 = 4 for all integers k . The same holds for the family x = 23 + πk . The
solutions which lie in [0, 2π ) come from the values k = 0 and k = 1, namely x = π , 23 , 43
and 3 . To conﬁrm graphically, we use a reciprocal identity to rewrite the secant as cosine.
The x-coordinates of the intersection points of y = (cos(x))2 and y = 4 verify our answers. y= cos(3x)
sin(3x) and y = 0 y= 1
cos2 (x) and y = 4 5. Our ﬁrst step in solving tan x = −3 is to consider values in the interval − π , π with a
tangent of −3. Since −3 isn’t among the ‘common values’ for tangent, we need the arctangent
function. The period of the tangent function is π , so we get x = arctan(−3)+ πk for integers k .
Multiplying through by 2 gives us our solution x = 2 arctan(−3)+2πk for integers k . To check
our answer, we note that for any integer k , tan 2 arctan(2 3)+2πk = tan (arctan(−3) + πk ) =
tan (arctan(−3)) = −3. To determine which of our answers lie in the interval [0, 2π ), we
begin substituting integers k into the expression x = 2 arctan(−3) + 2πk . When k = 0, we get
x = 2 ar...
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