# Solve the trigonometric equation

• Mar 5th 2009, 07:28 AM
james_bond
Solve the trigonometric equation
Solve for $x$, $y$: $25^{\sin^2x+2\cos y +1}+25^{\cos^2y+2\sin x +1}=2$.
• Mar 5th 2009, 07:42 AM
HallsofIvy
Quote:

Originally Posted by james_bond
Solve for $x$, $y$: $25^{\sin^2x+2\cos y +1}+25^{\cos^2y+2\sin x +1}=2$.

One possibility is that both those powers of 25 be equal to 1 so that 1+ 1= 2. In that case we must have $sin^2x+ 2 cos y+ 1= 0$ and $cos^2y+ 2\sin x+ 1= 0$. Adding those 2 equations, 1+ 2(cos x+ cos y)+ 2= 0 or cos x+ cos y= -3/2. From that cos y= -3/2- cos x. You can put that into $cos^2 x+ cos^2 y= 1$ to get a quadratic equation for cos x and then find x and y.
• Mar 5th 2009, 08:08 AM
red_dog
$\displaystyle 25^{\sin^2x+2\cos y+1}+25^{\cos^2y+2\sin x+1}\geq 2\sqrt{25^{(\sin x+1)^2+(\cos x+1)^2}}\geq 2$

The equality stands if $25^{\sin^2x+2\cos y+1}=25^{\cos^2y+2\sin x+1}=1\Rightarrow$

$\left\{\begin{array}{ll}\sin^2x+2\cos y+1=0\\\cos^2y+2\sin x+1=0\end{array}\right.$

$(\sin x+1)^2+(\cos y+1)^2=0\Rightarrow \sin x=\cos y=-1$