nunuskies
23.12.2019 •
Mathematics
Give the most general solutions to equations. 2sinxcosx-sin(2x)cos(2x)=0. a. simplify the first expression using double angle identity for sine.. b. factor the left side of equatioin. c. solve the factored equation
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Ответ:
Part I
The double angle identity for sine states that sin(2x) = 2sinxcosx
Thus we get:
sin(2x) - sin(2x)cos(2x) = 0
Part II
sin(2x)(1 - cos(2x)) = 0
Part III
Either sin(2x) = 0 or
1 - cos(2x) = 0
=> cos(2x) = 1
For sin(2x) = 0, this is true for
2x = n(pi) where n = 0, 1, 2,
x = n(pi/2)
For cos(2x) = 1, this is true for
2x = n(pi) where n = 0, 2, 4,
x = n(pi/2)
I hope my answer has come to your help. Thank you for posting your question here in .
Ответ:
iii) 5.7
iv) 27.85
vii) 0.609
viii) 40.0709
Step-by-step explanation:
hope this helps