4 sin(0) cos(0) + 2 cos(0) - 2 sin(0) - 1 = 0 Step 1 Begin by factoring the expression on the left. 4 sin(0) cos(0) + 2 cos(0) - 2 sin(0) - 1 = 0 (2 cos(0) - 1) 2 sin(0) + 1 Step 2 Now, set each factor equal to 0 and solve for 0. 2 sin(0) + 1 = 0 0 = 2 sin(0) = -1 sin(0) = 1 2 2 sin(0) + 1 ]) = ⁰ 0 1 2 2 cos(0) 1 = 0 2 cos(0) = 1 cos(0) = 2 1 2 Step 3 Because integer multiples of 2, added to the four values in the interval [0, 2) that meet this criteria, will give you all solutions for 0, it follows that for any integer k, we have the following.
4 sin(0) cos(0) + 2 cos(0) - 2 sin(0) - 1 = 0 Step 1 Begin by factoring the expression on the left. 4 sin(0) cos(0) + 2 cos(0) - 2 sin(0) - 1 = 0 (2 cos(0) - 1) 2 sin(0) + 1 Step 2 Now, set each factor equal to 0 and solve for 0. 2 sin(0) + 1 = 0 0 = 2 sin(0) = -1 sin(0) = 1 2 2 sin(0) + 1 ]) = ⁰ 0 1 2 2 cos(0) 1 = 0 2 cos(0) = 1 cos(0) = 2 1 2 Step 3 Because integer multiples of 2, added to the four values in the interval [0, 2) that meet this criteria, will give you all solutions for 0, it follows that for any integer k, we have the following.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:4 sin(0) cos(0) + 2 cos(0) - 2 sin(0) - 1 = 0
Step 1
Begin by factoring the expression on the left.
(2 cos(0) − 1)( 2 sin ( ✔) + 1
Step 2
Now, set each factor equal to 0 and solve for 0.
2 sin(0) + 1 = 0
4 sin(0) cos(0) + 2 cos(0) - 2 sin(0) - 1 = 0
2 sin(0)
0 =
sin (0)
=
=
-1
1
2
2 sin(0) +1
NI
= 0
2 cos(0) 1 = 0
2 cos(0)
cos(0)
=
1
1
2
1
Step 3
Because integer multiples of 2л, added to the four values in the interval [0, 2) that meet this criteria, will give you all solutions for 0, it follows that for any integer k, we have the
following.
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