Trigonometric equation.

Take the conjugate!! We will get sin x + √1+ sin² x = √1+ cos²x - cos x , Let solutions of this equations make a set [A] and negative of solutions of (- sin x) + √1+ sin² x = √1+ cos²x - cos x form a set [B]
It is easy to prove that they are subset of each other and hence are equal. So let us solve the other equation (- sin x) + √1+ sin² x = √1+ cos²x - cos x

We observe that the function -y + √1+ y² is strictly decreasing function. It's derivative is -1 + y√1+ y² which is less than zero for all y

So, we must have sin x = cos x, which means x = nπ + π4, nεZ

hence solutions of the original equation that is sin x + √1+ sin² x = √1+ cos²x - cos x is given by x = -( nÏ€ + Ï€4) , nεZ

Are you sure ?

For " sin x = cos x " , the given equation simply reduces to -

sin x + ( 1 + sin ² x ) ^{1 / 2} = 1 , - 1

Case 1 >

1 + sin ² x = 1 + sin ² x - 2 sin x

Or , sin x = 0 .

Case 2 >

1 + sin ² x = 1 + sin ² x + 2 sin x

Or , sin x = 0 .

But we had assumed " sin x = cos x " .

Hence , there are no solutions for " sin x = cos x " .

@ Ricky:Not so fast !! this is not the solution of original equation..negative of roots of sin x = cos x , are desired roots

Though you are correct , but this is much easier -

cos x + ( 1 + cos ² x ) ^{1 / 2} = ( 1 + sin ² x ) ^{1 / 2} - sin x .............. By inverting the " sin " part .

Similarly , cos x - ( 1 + cos ² x ) ^{1 / 2} = - ( 1 + sin ² x ) ^{1 / 2} - sin x ............... By inverting the " cos " part .

Addition gives us -

sin x = - cos x

Voila !