Periodic functions > The cosine function
123456The cosine function

Solutions to the exercises

Exercise 1
a

x 1,213 + k 2 π x -1,213 + k 2 π

b

x 1,928 + k 2 π x -1,928 + k 2 π

c

x = 1 6 π + k 2 π x = - 1 6 π + k 2 π

d

x = 3 4 π + k 2 π x = - 3 4 π + k 2 π

Exercise 2
a

x = k 2 π

b

x = 1 + k 2 π x = -1 + k 2 π

c

x = cos ( 1 ) 0,540

d

x 0,571 + k 2 π x -0,571 + k 2 π

Exercise 3
a

2 cos ( x ) - 1 = 0 gives cos ( x ) = 1 2 and so x = 1 3 π x = 1 2 3 π x = 2 1 3 π x = 3 2 3 π .
The zeroes are ( 1 3 π , 0 ) , ( 1 2 3 π , 0 ) , ( 2 1 3 π , 0 ) en ( 3 2 3 π , 0 ) .

b

1 3 π x 1 2 3 π 2 1 3 π x 3 2 3 π .

Exercise 4
a

cos ( 2 x ) = 0,5 gives 2 x = 1 3 π + k 2 π 2 x = - 1 3 π + k 2 π and so x = 1 6 π + k π x = - 1 6 π + k π .
On [ 0 , 4 π ] : x = 1 6 π x = 5 6 π x = 1 1 6 π x = 1 5 6 π x = 2 1 6 π x = 2 5 6 π x = 3 1 6 π x = 3 5 6 π .

b

0 x 1 6 π 5 6 π x 1 1 6 π 1 5 6 π x 2 1 6 π 2 5 6 π x 3 1 6 π 3 5 6 π x 4 π .

Exercise 5
a

sin ( x ) = 0 cos ( x ) = 0,5 gives x = 0 x = π x = 2 π x = 1 3 π x = 2 3 π .

b

sin 2 ( x ) - sin ( x ) = 0 gives sin ( x ) ( sin ( x ) - 1 ) = 0 and so sin ( x ) = 0 sin ( x ) = 1 .
Solutions: x = 0 x = π x = 2 π x = 1 2 π .

c

Using the quadratic formula you find cos ( x ) = 1 cos ( x ) = 0,5 and so x = 0 x = 2 π x = 1 3 π x = 1 2 3 π .

d

2 ( 1 - cos 2 ( x ) ) + cos ( x ) = 0 gives 2 cos 2 ( x ) - cos ( x ) - 2 = 0 and so cos ( x ) = 1 ± 9 4 so that cos ( x ) = 1 cos ( x ) = - 1 2 .
Solutions: x = 0 x = 2 π x = 2 3 π x = 1 1 3 π .

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