Solution to Trig Inequality in [0,-1]

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Homework Statement


Show that x=sin(piex) + cos(piex) has a solution in [0,-1]

The Attempt at a Solution



well i know that 0 <= x <= 1
therefore 0 <= sin(piex) + cos(piex) <= 1

But how do i go about solving this inequality?
 
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You don't go about solving it. You just prove it has a solution. Change the equation into f(x)=x-sin(pi*x)-cos(pi*x)=0. f(0)=(-1). f(1)=2. Does that suggest anything?
 
skateza said:

Homework Statement


Show that x=sin(piex) + cos(piex) has a solution in [0,-1]

The Attempt at a Solution



well i know that 0 <= x <= 1
therefore 0 <= sin(piex) + cos(piex) <= 1

But how do i go about solving this inequality?
That certainly is NOT true! What is sin(pi)+ cos(pi)? Look at the signs of sin(pix)+ cos(pix)- x at 0 and 1.
 
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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