Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 3 | A Maths
2 Answers Below
Anyone can contribute an answer, even non-tutors.
need help with this question. topic is trigonometric equations and identities
Date Posted: 3 years ago
Views: 321
cosec x = -k
1/sinx = -k
k = -1/sinx
Now,
-1 ≤ sinx ≤ 1 for all real values of x, as we always know.
Multiply by -1 and flip signs,
1 ≥ -sinx ≥ -1 → -1 ≤ -sinx ≤ 1
Now consider the inqualities separately.
-sinx ≥ -1 , -sinx ≤ 1
Take the reciprocal and flip the signs,
1/-sinx ≤ 1/-1 , 1/-sinx ≥ 1/1
- 1/sinx ≤ -1 , - 1/sinx ≥ 1
(When we take the reciprocal, the sign is flipped because the roles will change.
For example,
2 < 5 but ½ > 1/5
-½ < -⅓ but -2 > -3 )
Therefore, - 1/sinx ≤ -1 or - 1/sinx ≥ 1 and since k = - 1/sinx,
k ≤ -1 or k ≥ 1
So any value of k should be < 1 or > -1 such that -1 ≤ sinx ≤ 1 is not satisfied, giving no solution.
i.e -1 < k < 1
But,
k cannot be 0 since that would imply that -1/sinx = 0, which is not possible since we have a non-zero numerator , which means the fraction cannot be 0.
So -1 < k < 0 and 0 < k < 1
As long as you pick any value of k from this range, there will be solutions
1/sinx = -k
k = -1/sinx
Now,
-1 ≤ sinx ≤ 1 for all real values of x, as we always know.
Multiply by -1 and flip signs,
1 ≥ -sinx ≥ -1 → -1 ≤ -sinx ≤ 1
Now consider the inqualities separately.
-sinx ≥ -1 , -sinx ≤ 1
Take the reciprocal and flip the signs,
1/-sinx ≤ 1/-1 , 1/-sinx ≥ 1/1
- 1/sinx ≤ -1 , - 1/sinx ≥ 1
(When we take the reciprocal, the sign is flipped because the roles will change.
For example,
2 < 5 but ½ > 1/5
-½ < -⅓ but -2 > -3 )
Therefore, - 1/sinx ≤ -1 or - 1/sinx ≥ 1 and since k = - 1/sinx,
k ≤ -1 or k ≥ 1
So any value of k should be < 1 or > -1 such that -1 ≤ sinx ≤ 1 is not satisfied, giving no solution.
i.e -1 < k < 1
But,
k cannot be 0 since that would imply that -1/sinx = 0, which is not possible since we have a non-zero numerator , which means the fraction cannot be 0.
So -1 < k < 0 and 0 < k < 1
As long as you pick any value of k from this range, there will be solutions
See 2 Answers
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
As long as k <1, there will be no solution to the equation
Date Posted:
3 years ago
Actually for no solution,
-1 < k < 0 or 0 < k < 1
i.e 0< |k|< 1
k < 1 only will include values like k = -2
when k = -2, sin x = -1/-2 = ½ , which has a solution
-1 < k < 0 or 0 < k < 1
i.e 0< |k|< 1
k < 1 only will include values like k = -2
when k = -2, sin x = -1/-2 = ½ , which has a solution
Strictly speaking, it is not possible for k to be 0 since that results in 1/sinx = 0, which does not make sense since a non-zero numerator will never give a result of 0
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Hope you understand. Which part seems confusing for you? Is it the range of values of sinX part or solving the inequality with k? Or anything else?
Date Posted:
3 years ago
360 Tutoring Program