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secondary 3 | A Maths
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How to solve this?
Date Posted: 3 years ago
Views: 212
cos 3x = -0.5 for values of x from 0 to 180 (all my angles are in degrees)
0 < x < 180
is the same as
0 < 3x < 540
so in fact we need to solve cos 3x = -0.5 for values of “3x” from 0 to 540.
Basic reference acute angle, alpha
= cos inverse (0.5)
= 60 degrees
As you know it, 3x lies in the second and third quadrants.
Possible values of 3x
= 180 - alpha, 180 + alpha, (180 - alpha) + 360, … (the remaining angles are in fact out of range)
= 120, 240, 480
Possible values of x
= 40, 120, 160
0 < x < 180
is the same as
0 < 3x < 540
so in fact we need to solve cos 3x = -0.5 for values of “3x” from 0 to 540.
Basic reference acute angle, alpha
= cos inverse (0.5)
= 60 degrees
As you know it, 3x lies in the second and third quadrants.
Possible values of 3x
= 180 - alpha, 180 + alpha, (180 - alpha) + 360, … (the remaining angles are in fact out of range)
= 120, 240, 480
Possible values of x
= 40, 120, 160
Make sure you scale the angle range according to the appropriate range (and not just take 0 to 180).
See 2 Answers
Thx for helping me! But there are more angles cos it's 0 to 180 degree (I think) I wish to know how to get all the other angles :( but again, thank you for your help
the other angle is 320, out of range
Oh ic.. But the other two are 80 and 160 according to the ans key (maybe it's wrong
oh, sry, i understyod wrongly.
then the ans are 40, 80 and 160, it would be easier to see if you draw the graph
then the ans are 40, 80 and 160, it would be easier to see if you draw the graph
done
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hope its easier to see
Date Posted:
3 years ago
Ohh thx!! I got it
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