Solve the trigonometric equation

**james_bond** · March 5th 2009, 07:28 AM

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

**HallsofIvy** · March 5th 2009, 07:42 AM

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.

**red_dog** · March 5th 2009, 08:08 AM

$\displaystyle 25^{\sin^2x+2\cos y+1}+25^{\cos^2y+2\sin x+1}\geq 2\sqrt{25^{(\sin x+1)^2+(\cos x+1)^2}}\geq 2$

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

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

Adding the equations we have

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

Now find x and y.

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