cosec x - sin x = 2 for all values in the interval 0< x < Pi?

Question

Find the value of x please.

I tried to solve this by using the fact cosec x is 1/sin x

therefore 1/sin x - sin = 2

The brining two to the otherside and multiplying through by sin x

getting 1-sin (squared) x -2.

I got stuck from here on!!

Answer 1

csc x=1/sin x so
1/sin x-sin x=2 multiply both sides by sin x
1-sin^2 x=2 sin x subtract 2 sin x from both sides
-sin^2 x -2*sin x+1=0 multiply by -1
sin^2 x+2sin x-1=0

sin x=(-2+/-√4+4))/2
sin x=-1+/-√2
sin x=-1-√2 has no solution since -1<=sin x<=+1
sin x=-1+√x=.41421
x=arcsin 0.412.7145
x=0.4271 radians in the 1st quadrant &
x= π-0.4271=2.7145 radians in the 2nd

Answer 2

csc(x) - sin(x) = 2, for 0 < x < pi

Your first step is to convert everything to sines and cosines.

1/sin(x) - sin(x) = 2

Now, move the 2 over to the left hand side.

1/sin(x) - sin(x) - 2 = 0

Put them all under a common denominator.

1/sin(x) - sin^2(x)/sin(x) - 2sin(x)/sin(x) = 0

Merge the three fractions into 1.

[1 - sin^2(x) - 2sin(x)] / sin(x) = 0

Let's multiply both sides by (-1),

[-1 + sin^2(x) + 2sin(x)] / sin(x) = 0

And now, let's put the sines in descending power.

[sin^2(x) + 2sin(x) - 1] / sin(x) = 0

Normally, we cannot always do this next step, but given that x = 0 is not a solution to this problem, we can multiply both sides by
sin(x), and that will give us

sin^2(x) + 2sin(x) - 1 = 0

And now, we solve this like it were a quadratic.

sin(x) = [-2 +/- sqrt(4 - 4(-1))] / 2
sin(x) = [-2 +/- sqrt(8)] / 2
sin(x) = [-2 +/- 2sqrt(2)] / 2
sin(x) = -2 +/- sqrt(2)

So we have two equations to work with:

sin(x) = -2 + sqrt(2) and
sin(x) = -2 - sqrt(2)

Note that we can discard the second equation, since -2 - sqrt(2) can never occur (as -2 - sqrt(2) approximates to -3.41, and
-1 <= sin(x) <= 1). That means

sin(x) = -2 + sqrt(2)

[-2 + sqrt(2)] is a negative number, and sine is negatively only in quadrants 3 and 4. However, if x is in the interval 0 to pi, then that suggests x is in quadrants 1 and 2. This cannot happen.

Therefore, there is no solution.

Answer 3

You should get : 1 - sin^2x - 2sinx = 0
Rearrange and solve this quadratic for sinx:

sin^2x + 2sinx -1 = 0
The roots are sinx = (-2 +- sqrt(2^2 - 4*1*(-1)))/2*1
= -1 +- sqrt(2)
Now, with -sqrt(2), sinx will be less than -1. Not possible. So use the other root

Sinx = -1 + sqrt(2) = 0.414
x = sininv(0.414)

Answer 4

cosec x - sin x = 2, 0
1/sinx-sinx = 2
multiply by sinx
1-sin^2=2sinx
sin^2(x)+2sinx-1=0
{sinx+(1-sqrt2)}
{sinx+(1+sqrt2)}=0
>>>sinx= -1 +/- sqrt2
-1 hence,sinx=sqrt(2)-1

x=arcsin(sqrt(2)-1)
=24.46980052 degrees
and (180- 24.46980052)
=155.5301995 degrees

{sin is +ve in 1st and 2nd
quadrants}

i hope that this helps

Answer 5

Let u = sin x.
1/u - u = 2

Multiply both sides by u,
u^2+2u-1 = 0

Use quadratic formula with |u|≦1,
u = .414
x = arcsin u = .427
----
Check:
LHS = 1/csc(.427) + sin(.427) = 2.0000 = RHS

Answer 6

ƒ(x) = cos x - sin x = √2.[ (cos x)(1/√2) - (sin x)(1/√2) ] . . . .= √2. [ cos x. cos π/4 - sin x. sin π/4 ] . . . .= √2. cos (x + π/4) Now, for all real x, ... -1 ≤ cos (x + π/4) ≤ 1 ∴ -√2 ≤ √2. cos (x + π/4) ≤ √2 ∴ -√2 ≤ ƒ(x) ≤ √2 ∴ ƒmin = -√2, ... ƒmax = √2 ............. Ans. __________________________________ Happy To Help ! ___________________________________

Answer 7

pi=3.14