Question #d5864

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Question #d5864

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Answer 1 · 2017-05-19T04:46:54

pi/8; (3pi)/8; (5pi)/8; (7pi)/8; (9pi)/8; (11pi)/8; (13pi)/8; (15pi)/8

Explanation:

cos 4x = 0
The unit circle gives 2 solutions:
$4 x = \frac{π}{2}$ $4x = pi/2$ , and $4 x = \frac{3 π}{2}$ $4x = (3pi)/2$ .
a. $4 x = \frac{π}{2} + 2 k π$ $4x = pi/2 + 2kpi$
$x = \frac{π}{8} + \frac{k π}{2}$ $x = pi/8 + (kpi)/2$
b. $4 x = \frac{3 π}{2} + 2 k π$ $4x = (3pi)/2 + 2kpi$
$x = \frac{3 π}{8} + \frac{k π}{2}$ $x = (3pi)/8 + (kpi)/2$
8 answers for $[0, 2 π]$ $[0, 2pi]$ --> k = 0, 1, 2, 3,
k = 0 --> $x = \frac{π}{8}$ $x = pi/8$ ;
k = 1 --> $x = \frac{π}{8} + \frac{π}{2} = \frac{5 π}{8}$ $x = pi/8 + pi/2 = (5pi)/8$ ;
k = 2 --> $x = \frac{π}{8} + π = \frac{9 π}{8}$ $x = pi/8 + pi = (9pi)/8$ ,
k = 3 --> $x = \frac{π}{8} + \frac{3 π}{2} = \frac{13 π}{8}$ $x = pi/8 + (3pi)/2 = (13pi)/8$
k = 0 --> $x = \frac{3 π}{8}$ $x = (3pi)/8$ ;
k = 1 --> $x = \frac{3 π}{8} + \frac{π}{2} = \frac{7 π}{8}$ $x = (3pi)/8 + pi/2 = (7pi)/8$
k = 2 --> $x = \frac{3 π}{8} + π = \frac{11 π}{8}$ $x = (3pi)/8 + pi = (11pi)/8$ ;
k = 3 --> $x = \frac{3 π}{8} + \frac{3 π}{2} = \frac{15 π}{8}$ $x = (3pi)/8 + (3pi)/2 = (15pi)/8$

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Question #d5864

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