173. An interesting trigonometric equations.
by Virgil Nicula, Nov 20, 2010, 9:04 AM
Prove that the trigonometrical equation 
, where 
hasn't solutions.
Proof. Observe that
.
Therefore,
. Since 
obtain that 
.
Remark. From the above proof hold the identity
. Using this relation for 
can obtain easily that in any triangle
exists the inequalitity 
, where 
. Indeed,
and 
a.s.o.
PP1. Ascertain
for which the trigonometrical equation 
has at least one solution in ![$\left[0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/2/1/8/218957f184ff318bffc0781452c79af681ced9a0.png)
.
Proof. Consider the function
, where ![$E=\left[0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/7/3/1/7317ac347579f3b05602f3643e3bea4621f40852.png)
and 
. Then 
Observe that
and for any 
have ![$\boxed{\sin x+\cos x=t}\in\left[1,\sqrt 2\right]$](//latex.artofproblemsolving.com/f/e/a/fea48095fd5c6854d2cc6a2c16c7e959ff5a4133.png)
. Thus, 
. Denote 
, where ![$t\in F=\left[1,\sqrt 2\right]$](//latex.artofproblemsolving.com/5/6/f/56f2b324df31044443c19e0d1bc4acccc3601ee5.png)
. In conclusion, ![$\lambda\in f(E)=g(F)=\left[\sqrt 2-1,1\right]$](//latex.artofproblemsolving.com/8/b/3/8b338eb483789cb489e6f9668712ad46b7bd2753.png)
.
Remark. For
and for any 
we have 
. Otherwise, 
if and only if
, what is truly. We"ll have equality if and only if 
.
PP2. Solve the trigonometrical equation
.
Proof. Must
and our equation is equivalently with 
![$\cos 2x\left[\sin\left (x+\frac{\pi}3\right)-1\right]=0$](//latex.artofproblemsolving.com/d/9/c/d9cc0ecec47ffab8805d33891f12d060d448be8d.png)
. Hence the solution is 
.
, where 
hasn't solutions.Proof. Observe that
.Therefore,
. Since 
obtain that 
.Remark. From the above proof hold the identity
. Using this relation for 
can obtain easily that in any triangle
exists the inequalitity 
, where 
. Indeed,
and 
a.s.o.PP1. Ascertain
for which the trigonometrical equation 
has at least one solution in ![$\left[0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/2/1/8/218957f184ff318bffc0781452c79af681ced9a0.png)
.Proof. Consider the function
, where ![$E=\left[0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/7/3/1/7317ac347579f3b05602f3643e3bea4621f40852.png)
and 
. Then 
Observe that
and for any 
have ![$\boxed{\sin x+\cos x=t}\in\left[1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/f/e/a/fea48095fd5c6854d2cc6a2c16c7e959ff5a4133.png)
. Thus, 
. Denote 
, where ![$t\in F=\left[1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/5/6/f/56f2b324df31044443c19e0d1bc4acccc3601ee5.png)
. In conclusion, ![$\lambda\in f(E)=g(F)=\left[\sqrt 2-1,1\right]$](http://latex.artofproblemsolving.com/8/b/3/8b338eb483789cb489e6f9668712ad46b7bd2753.png)
.Remark. For
and for any 
we have 
. Otherwise, 
if and only if
, what is truly. We"ll have equality if and only if 
.PP2. Solve the trigonometrical equation
.Proof. Must
and our equation is equivalently with 
![$\cos 2x\left[\sin\left (x+\frac{\pi}3\right)-1\right]=0$](http://latex.artofproblemsolving.com/d/9/c/d9cc0ecec47ffab8805d33891f12d060d448be8d.png)
. Hence the solution is 
.This post has been edited 31 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:16 AM
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