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.
How to solve this trigonometric equation?
-
0The [cosine expansion formula](http://www.wolframalpha.com/input/?i=cos(4x)) might help. – 2012-10-22
-
0Since solving the equation seems to be difficult, why not tell us how the equation arose. Maybe there's a way around it. – 2012-10-22
2 Answers
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.
-
0And then what do you do with the $\cos4x$? – 2012-10-22
-
0You 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. – 2012-10-22
-
0@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 $\sin y = \frac{1}{\sqrt 2}$). – 2012-10-22
-
0@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 $0$. – 2012-10-22
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
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 $x = \pi + 2m\pi.$
-
0@minthano_2011 : You solution improves on mine. – 2012-10-22
-
0How do you justify that the only solution of that 5th degree equation is $t=1$? – 2012-10-23
-
0@GerryMyerson : see minthao_2011's solution above. $f'(t) = 5t^4 - 6t^2 + 2 > 0$ for all real $t$. – 2012-10-23