I want to solve the equation $$(3-\cos 4x)(\sin x - \cos x)=2.$$ I solve by putting $t = \sin x - \cos x$, but I can not find all solutions.
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$\begingroup$ The cosine expansion formula might help. $\endgroup$– BenOct 22, 2012 at 1:15
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$\begingroup$ Since solving the equation seems to be difficult, why not tell us how the equation arose. Maybe there's a way around it. $\endgroup$– Gerry MyersonOct 22, 2012 at 5:03
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2 Answers
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I have just a solution. \begin{equation*} (3-\cos 4x)(\sin x - \cos x) = 2. \end{equation*}
Note that, $3-\cos 4x > 0$, therefore $\sin x - \cos x>0$.
Put $t = \sin x - \cos x = \sqrt{2}\sin\left(x - \dfrac{\pi}{4}\right)$, and then $0<t \leqslant \sqrt{2}$. We have $$ t^2 = (\sin x - \cos x)^2 = 1 -\sin 2x.$$ Another way, $$3 - \cos 4x =3 - (1-2\sin^2 2x )= 2[1-(1-t^2)^2] = 2(t^4 -2t^2 + 2)$$ The given equation has the form $$2(t^4 -2t^2 + 2)t = 2,$$ equavalent to $$t^5- 2t^3+2t - 1 = 0.$$ Now we prove that this equation has only root $t = 1$.
First way. The function $f(t)=t^5 -2t^3 +2t - 1$ has $$f'(t) = 5t^4-6t^2+2>0, \forall t$$ Therefore, $f$ is an increasing function on the interval $(0; \sqrt{2}]$. And $f(1) = 0$, thus $t = 1$ is only root.
Second way, we have $$t^5- 2t^3+2t - 1 = 0 \Leftrightarrow (t-1)(t^4 +t^3 -t^2 -t + 1)=0.$$ Note that $$t^4 +t^3 -t^2 -t + 1 = \biggl(t^2 + \dfrac{t}2 -1\biggr)^2 + \dfrac{3t^2}4 > 0, \forall t.$$ Thirt way, we have $$t^4 +t^3 -t^2 -t + 1=0 \Leftrightarrow \biggl(t - \dfrac{1}{t}\biggr)^2 + \biggl(t - \dfrac{1}{t}\biggr) + 1 = 0. $$ The last equation has no sulution. With $t = 1$, we have $\sin x - \cos x = 1$, then $x = \dfrac{\pi}2 + k2\pi$ and $x = \pi + 2m\pi.$
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$\begingroup$ @minthano_2011 : You solution improves on mine. $\endgroup$ Oct 22, 2012 at 13:16
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$\begingroup$ How do you justify that the only solution of that 5th degree equation is $t=1$? $\endgroup$ Oct 23, 2012 at 12:26
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$\begingroup$ @GerryMyerson : see minthao_2011's solution above. $f'(t) = 5t^4 - 6t^2 + 2 > 0$ for all real $t$. $\endgroup$ Oct 23, 2012 at 14:02
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Use $\sin x - \cos x = \sqrt{2}(\sin(x - \frac{\pi}{4}))$ followed by the substitution $x = y + \frac{\pi}{4}$. You get $(3+ \cos(4y))\sin y = \sqrt{2}$. This is satisfied if $\sin y = \frac{1}{\sqrt{2}}$. The equation $(3+ \cos(4y))\sin y = \sqrt{2}$ can be written as a fifth-degree polynomial in $\sin y$, and you know one of the roots, so you can get a fourth-degree polynomial.
answered Oct 22, 2012 at 0:35
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$\begingroup$ And then what do you do with the $\cos4x$? $\endgroup$ Oct 22, 2012 at 0:40
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$\begingroup$ You substitute $x$ with $y + \frac{\pi}{4}$, as I said. I must apologize to the OP - I posted in haste and did not notice that this does not completely solve the problem. But it simplifies it. You will get a $\cos(4y)$, which can be expressed as a fourth-degree polynomial in $\sin y$. So the original equation can be transformed to a fifth-degree polynomial in $\sin y$. I can't think of anything better at the moment. $\endgroup$ Oct 22, 2012 at 0:44
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$\begingroup$ @minthao_2011 : I plotted $(3+\cos(4y))*\sin(y)-\sqrt{2}$ from $0$ to $2\pi$ (see my answer above) in Maple and the only zeros were at $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ (where $\sin y = \frac{1}{\sqrt 2}$). $\endgroup$ Oct 22, 2012 at 1:25
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$\begingroup$ @minthao_2011 : See my answer above. Setting $Y = \sin y$, you get $4Y^4 - 2\sqrt{2}Y^3 - 2Y^2 -2\sqrt{2}Y + 1 = 0$. I plotted the left hand side of the equation in Maple from $Y = -1$ to $Y = 1$ and it was always positive, but unfortunately it got quite close to $0$. $\endgroup$ Oct 22, 2012 at 1:50