SOLUTION: Solve: Sqrt(sin(x) - sqrt(sin(x) + cos(x))) = cos(x)

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Question 1160739: Solve:
Sqrt(sin(x) - sqrt(sin(x) + cos(x))) = cos(x)
Found 2 solutions by KMST, ikleyn:
Answer by KMST(5422) About Me  (Show Source):
You can put this solution on YOUR website!
GRAPHING CALCULATOR PROBLEM?
This may be one of those questions designed for using one of those graphing calculators,
that give you solutions to equations without requiring reasoning and understanding.
In this website, I can easily graph too, and at least find the approximate solution or solutions that way.
If the approximate solutions found are suspiciously close to something like x=0 or x=pi%2F2 ,
we can verify if they are exactly that, and may find exact solutions that way.
However, that does not require much understanding or reasoning,
and is a sad outsourcing to a device of the reasoning that makes humans unique.

ANOTHER WAY?
Maybe a graphing calculator is not needed.
Maybe there is even an elegant simple and obvious solution that we cannot quite see right now.
Let us see how we can use some reasoning and understanding.

Do we need to know a lot about the functions sin%28x%29 and cos%28x%29 ?
Maybe all we need to know is not much more than that cos%5E2%28x%29%2Bsin%5E2%28x%29=1.
Maybe we can replace the expressions in the original equation with something simpler.

How about a=cos%28x%29 and b=sin%28x%29 with the condition that a%5E2%2Bb%5E2=1 ?
a%5E2%2Bb%5E2=1 means that we cannot have a=b=0 .
We could have a=0 or b=0 , but not both.
When one is zero, the square of the other must be 1.
a%5E2%2Bb%5E2=1 also means that a%5E2%3C=1 and b%5E2%3C=1

With that change of variable the problem simplifies to
sqrt%28b-sqrt%28a%2Bb%29%29=a and that looks easier to handle.
I will color the inside of the first square root, to make it easier to track those nested square roots:
sqrt%28red%28b-sqrt%28a%2Bb%29%29%29=a

We know that sqrt%28red%28b-sqrt%28a%2Bb%29%29%29%3E=0 if it exists, so that means Highlight%28a%3E=0%29
and that for sqrt%28red%28b-sqrt%28a%2Bb%29%29%29 to exist, we must have
red%28a%2Bb%29%3E=0 and red%28b-sqrt%28a%2Bb%29%29%3E=0

If they exist, red%28sqrt%28a%2Bb%29%29%3E=0 and
red%28b-sqrt%28a%2Bb%29%29%3E=0 --> b%3E=sqrt%28a%2Bb%29%3E=0 --> b%3E=0 .
Now we know that it must be 0%3C=a%3C=1 , 0%3C=b%3C=1 and with a%5E2%2Bb%5E2=1 it cannot be a=b=0

Could it be that a%3E0 and b%3E0 ?
In that case,
ab%3E0 --> 2ab%3E0 --> %28a%2Bb%29%5E2=a%5E2%2Bb%5E2%2B2ab=1%2B2ab%3E1 --> a%2Bb%3E1 --> sqrt%28a%2Bb%29%3E1 --> b-sqrt%28a%2Bb%29%3C0

Then, it must be either a=0 with b=1 or b=0 with a=1 .

Can we have red%28b-sqrt%28a%2Bb%29%29%3E=0 and sqrt%28red%28b-sqrt%28a%2Bb%29%29%29=a in either case?

If {{b=0}}} with a=1 , we get red%28b-sqrt%28a%2Bb%29%29=red%280-sqrt%281%2B0%29%29=red%28-1%29 , so that cannot give us a solution.
Then it must be a=0 and b=1 , and if b=1=sin%28x%29 ,
then highlight%28x=pi%2F2%29 or highlight%28x=pi%2F2%2Bn%2A2pi%29 where n is an integer.

Answer by ikleyn(53996) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve:
Sqrt(sin(x) - sqrt(sin(x) + cos(x))) = cos(x)
~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Our starting equation is 

    sqrt%28sin%28x%29+-+sqrt%28sin%28x%29+%2B+cos%28x%29%29%29 = cos(x).    (1)


Let's consider function  sin%28x%29+-+sqrt%28sin%28x%29+%2B+cos%28x%29%29  first, which is under the outer square root of the left side.


Inside the first quadrant, QI, when  0 < x < pi%2F2,  both sin(x) and cos(x) are positive.

Therefore, the sum sin(x) + cos(x) is greater than sin(x): 

    sin(x) + cos(x) > sin(x).     (2)


Square root is a monotonic function of its argument; therefore, from inequality (1) we have

    sqrt%28sin%28x%29+%2B+cos%28x%29%29 > sqrt%28sin%28x%29%29.    (3)


Since inside the first quadrant, QI, sin(x) is less than 1, we have

    sqrt%28sin%28x%29%29 > sin(x).   (4)


So, combining (3) and (4), we have

    sin(x) < sqrt%28sin%28x%29+%2B+cos%28x%29%29.


It means that inside QI the expression under the outer square root is negative;
so, left side of the given equation is not defined.


Let's inspect endpoints of QI, angles x = 0 and x = pi%2F2.


At x = 0, left side of the given equation is not defined.

At x = pi%2F2, the given equation is valid, so x = pi%2F2 is a solution.


Thus we found that in the first quadrant, 0 <= x <= pi%2F2, the only solution is  x = pi%2F2,   
where both sides of the given equation are defined and the equation is valid.



In the second quadrant,  pi%2F2 < x <= pi,  the left side of the given equation 

is EITHER positive OR not defined, while the right side is always negative, so in the domain

pi%2F2 < x <= pi our given equation has no solutions.



In QIII and QIV, where  pi < x < 2pi,  

    (a) sin(x) is negative 

and the term

    (b) -sqrt%28sin%28x%29%2Bcos%28x%29%29  is either negative or not defined.


Therefore, the left side of the given equation is not defined at  pi < x  < 2pi.


So, in the union  { QIII U QIV ) there is no solution to equation (1), at all.


Thus we found that the only solution to equation (1) in the interval  [0,2pi)  is  x = pi%2F2.


If you want to get the GENERAL solution for any real x, then use the fact that both
left side and right side of equation (1) are periodic functions of x with the period 2pi.


Hence, the general solution to equation (1) is the set of values  pi%2F2+%2B+2k%2Api,  
where  'k'  is any integer  k = 0, +/-1, +/-2, . . .  and so on.


At this point, the problem is solved completely.

The method of solution is an accurate analysis of left side and right side of the equation.