Maxima and Minima of tan(sinx + sin3x) on (-pi,pi)

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Homework Statement


Find the maxima and minima of the function f(x)=tan(sinx + sin3x) on the interval (-pi,pi)


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The Attempt at a Solution


I found the derivative as sec^2x (sinx +sin3x) (cosx + cos3x) and when i set it equal to zero i found critical points at Sin=pi/2, and cos=pi/2
 
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Your derivative is incorrect. The derivative of sin(3x) is not cos(3x)...
 
Wait i typed that part wrong. i had derivative of sin(3x) as 3Cos(3x)
 
Good, so your derivative is

\frac{\cos(x)+3\cos(3x)}{\cos^2(\sin(x)+\sin(3x))}

This can only become zero if \cos(x)+3\cos(3x). So you'll need to solve the equation \cos(x)=-3\cos(3x). I believe there will be infinitly many solutions to this, but I'm not sure...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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