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So I have $\sin^4x+\cos^4x=m$ and $\sin^6x+\cos^6x=m$, and I need to find $m\geq0$ such that the real solutions of these two equations to be common.

I started by changing the form of these equations and I got $$\frac{1}{2}\leq 1-\frac{\sin^{2}(2x)}{2}\leq 1$$ for first equation and $$\frac{1}{4}\leq 1-\frac{3\sin^{2}(2x)}{4}\leq 1$$ for the second equation.

So I find the values of m such that these equation has solution and I think that to have common solutions these intervals must be intersected and I got m from $[1/2,1]$, but the right answer is $m=1$.

How to approach this exercise?

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Hint: Use that $$\sin(x)^4+\cos(x)^4=\frac{1}{4}(3+\cos(4x))$$ and $$\sin(x)^6+\cos(x)^6=\frac{1}{8}(5+3\cos(4x))$$

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You may proceed as follows by first finding all $m$ such that $$\sin^4x+\cos^4x= \sin^6x+\cos^6x = m$$

To do so,

  • set $u = \sin^2 x, v= \cos^2 x$
  • $\Rightarrow u+v=1$ and $u^3+v^3 = u^2+v^2 = m$

Now you get $$\Rightarrow 1 = (u+v)^2 = m +2uv \Rightarrow uv = \frac{1-m}{2}$$ $$\Rightarrow 1 = (u+v)^3 = m +3uv(u+v) \Rightarrow m + \frac{3}{2}(1-m) = 1 \Rightarrow \boxed{m=1}$$

Obviously, for $m=1$ both equations have the same solutions.

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I would try to replace $sin^2(x)$ with $1-cos^2(x)$ in the equation

$$sin^4(x) + cos^4(x) = sin^6(x) + cos^6(x) $$

You can work out $sin(x)$ from there and then get $m$

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