Fr what values of the constant k does sinx + cosx = k have exactly two solutions in the interval 0<x<360

Question

what has the constant k got to do with anything!? please help...working is great and any explanations are welcome! ta

Answer 1

Try sketching the graph of y = sinx + cosx. A graphic calculator might help.

You will notice the graph ranges from √2 to - √2.

Some points that lie on the graph:
(0,1), (&pi/4, √2), (&pi/2, 1), (3&pi/4,0), (&pi, -1), (5&pi/4, -√2), (3&pi/2, -1), (7&pi/4, 0), (2&pi, 1)

[I prefer using radians, the degrees are too messy... If you like, just write 180 degrees instead of &pi]

Now draw horizontal lines across the graph. If you draw a line at y=√2, it will cut the graph exactly once. Same for y=-√2.

Now the question is whether you are to find the solutions in the interval
0 or
0 <= x <= 360
(i.e., are 0 and 360 included or not)

Anyway; if they are included, the line y=1 will cut the graph exactly 3 times, but if they are not, then only once. So it doesn't really matter.

Every other line y=k will cut the graph exactly 2 times. So the solutions for k are all the numbers in-between -√2 and √2 except for 1.

We can write this as a union of 2 intervals:
k belongs to (-√2, 1) U (1, √2)

Or as one interval with value {1} excluded:
k belongs to (-√2, √2) - {1}

Or as inequalities:
-√2 < k < 1 or 1 < k < √2

(whichever way you give the answer, it must be obvious that -√2 < k < √2, but k cannot be 1)

I hope my explanation is not too difficult to understand. You can contact me for more info if needed.

Answer 2

Working with tan(x/2)=z
2z+1-z^2=k(1+z^2)
(1+k)z^2 -2z +k-1=0

4-4((k+1)(k-1) >0 so 1-k^2+1>0 so k^2<2 and
-sqrtr2 < k

Answer 3

let us convert it to the form r sin (x+/-t)

let 1(coefficient of sin x) = r cos t
1 (coefficient of cos x) = r sin t

squaire and add r = sqrt(2)

sin t = 1/ sqrt(2)

t = pi/4

so we get sqrt(2)( sin x cos pi/4 + cos x sin pi/4) = k
or sqrt(2) sin (x+pi/4) = k

now what is the range of sin x that is from -1 to + 1

so krange is -sqrt(2) to + sqrt (2)

now we know the impliction of k

if k is < - sqrt(2) no solution
if k > sqrt(2) no solution
if k = sqrt(2) sin (x+pi/4) = 1 so one solution x+pi/4 = pi/2 or x = pi/4
if k = - sqrt(2) sin (x+pi/4) = -1 so one solution
x+pi/4 = 3pi/2 or x = 5pi/4


but if -sqrt(2) < k < sqrt(2) then 2 solutions