How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$

Question

How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$

I don't know the solution for this. Help me!

Thank all!

Answer 1

The use of the "multiple-angle" formulas is probably the surest way to establish the solutions to this equation; however, it requires a goodly amount of algebraic manipulations. We can also say something about the solutions by investigating the properties of the terms in the equation

$$( \ \sin x \cdot \sin 2x \cdot \sin 3x \ ) \ + \ ( \ \cos x \cdot \cos 2x \cdot \cos 3x \ ) \ = \ 1 \ .$$

Something to consider is that both product terms of the sum have the same period, since the individual factors have periods of $ \ 2 \pi \ , \ \pi \ , \ \text{and} \ \frac{2 \pi}{3} \ . $ The period of each product term is $ \ \pi \ $ , since both $ \ \sin x \ \ \text{and} \ \ \sin 3x \ $ (or $ \ \ \cos x \ \ \text{and} \ \ \cos 3x \ $ ) complete an odd number of half-cycles in intervals of that length, restoring each product to the same parity at $ \ x = \pi \ $ that it had at $ \ x = 0 \ . $ So we only need to focus for the present on the behavior of the terms in the interval $ \ [ \ 0 \ , \ \pi \ ) \ . $

The first term has zeroes at all multiples of $ \frac{\pi}{2} \ \text{and} \ \frac{\pi}{3} \ $ , while the second term has its zeroes at all odd multiples of $ \frac{\pi}{6} \ , \ \frac{\pi}{4} \ \text{and} \ \frac{\pi}{2} \ . $ Upon checking the sign of each factor in both product terms, we find that the "product of sines" is positive on $ \ ( \ 0 \ , \ \frac{\pi}{3} \ ) \ \ \text{and} \ \ ( \ \frac{\pi}{2} \ , \ \frac{2 \pi}{3} \ ) \ , $ while the "product of cosines" is positive on $ \ ( \ 0 \ , \ \frac{\pi}{6} \ ) \ , \ ( \ \frac{\pi}{4} \ , \ \frac{\pi}{2} \ ) \ , \ ( \ \frac{\pi}{2} \ , \ \frac{3 \pi}{4} \ ) \ , \ \text{and} \ \ ( \ \frac{5 \pi}{6} \ , \ \pi \ ) \ . $ Both terms, then, are positive only on $ \ ( \ 0 \ , \ \frac{\pi}{6} \ ) \ , \ ( \ \frac{\pi}{4} \ , \ \frac{\pi}{3} \ ) \ , \ \text{and} \ \ ( \ \frac{\pi}{2} \ , \ \frac{2 \pi}{3} \ ) \ . \ $ [There are, of course, corresponding "windows" in every other periodic interval.] This is important to what follows.

As the terms are products of sine and cosine functions, it is the case that

$$-1 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ +1 \ , $$

and similarly for the product-of-cosines. But we can be more specific about the range of these terms in the intervals where both are positive.

At $ \ x = 0 \ , $ $ \ \cos x \cdot \cos 2x \cdot \cos 3x \ $ plainly equals 1 and falls to zero at $ \ x = \frac{\pi}{6} \ $ , since all three factors are decreasing. In the same interval, all three factors of $ \ \sin x \cdot \sin 2x \cdot \sin 3x \ $ are increasing and

$$0 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ (\sin \frac{\pi}{6}) \cdot (\sin \frac{\pi}{3}) \cdot (\sin \frac{\pi}{2}) \ \le \ \frac{\sqrt{3}}{4} \ \approx \ 0.433 \ . $$

So we can be reasonably satisfied that the only solution to the equation in the interval $ \ [ \ 0 \ , \ \frac{\pi}{6} \ ] \ $ is $ \ x = 0 \ . \ $ [It may require a bit more work -- short of making a graph -- to be completely satisfied that this is so.]

In the interval $ \ ( \ \frac{\pi}{4} \ , \ \frac{\pi}{3} \ ) \ , \ $ we note that

$$0 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ (\sin \frac{\pi}{3}) \cdot (\sin \frac{\pi}{2}) \cdot (\sin \frac{3 \pi}{4}) \ \le \ \frac{\sqrt{6}}{4} \ \approx \ 0.612 \ $$

and

$$0 \ \le \ \cos x \cdot \cos 2x \cdot \cos 3x \ \le \ (\cos \frac{\pi}{4}) \cdot (\cos \frac{2 \pi}{3}) \cdot (\cos \pi) \ \le \ \frac{\sqrt{2}}{4} \ \approx \ 0.354 \ . $$

Thus, there is no prospect that the sum of these terms can equal 1 in this interval. (The fact that the individual terms change in opposite directions over the interval guarantees this. And, in any event, the products do not even attain the indicated upper bounds.)

Finally, on $ \ ( \ \frac{\pi}{2} \ , \ \frac{2 \pi}{3} \ ) \ , \ $ we may conclude that

$$0 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ (\sin \frac{\pi}{2}) \cdot (\sin \frac{4 \pi}{3}) \cdot (\sin \frac{3 \pi}{2}) \ \le \ \frac{\sqrt{3}}{2} \ \approx \ 0.866 \ $$

and

$$0 \ \le \ \cos x \cdot \cos 2x \cdot \cos 3x \ \le \ (\cos \frac{2 \pi}{3}) \cdot (\cos \frac{4 \pi}{3}) \cdot (\cos 2 \pi) \ \le \ \frac{1}{4} \ . $$

Somewhat closer checking is needed to show that the "product of sines" falls well short of the upper bound suggested here, and that this product reaches its maximum and then decreases before the "product of cosines" attains its upper bound. So, again, the sum of terms fails to reach 1 in this interval.

Therefore, in the entire interval $ \ [ \ 0 \ , \ \pi \ ) \ , \ $ the only solution found for the equation is $ \ x = 0 \ . $ This means that the set of all solutions to the equation contains only all integer multiples of $ \ \pi \ . $ A graph of each term and of their sum is presented below.

The orange curve represents the "product of sines", the green curve, the "product of cosines", and the blue curve, their sum.

Answer 2

The question can be solved with the help of calculus.

$f(x) = \sin x\sin2x\sin 3x + \cos x\cos 2x\cos 3x$

$\implies (1/2)(\cos x -\cos3x)\sin3x +(1/2)(\cos3x +\cos x)\cos3x.$

Simplifying once more,

$\implies (1/4)(\sin4x - \sin2x - \sin6x + 1+\cos4x +\cos4x+ \cos x).$

Finding the maximum of $f(x)$ gives $1$ for $x =0$ (by inspection).

Also the period of the function is LCM of $\pi$, $\pi/2$ , & $\pi/3$ which is $\pi$.

Therefore, the general solution of $f$ is

$n\pi$ for $n\in I$.

Answer 3

Let $\mathbf v(x) = (\sin x \cdot \sin 2x, \cos x \cdot \cos 2x) \in \mathbb R^2$. Compute:

$$\begin{align} |v|^2 &= \sin^2 x \cdot \sin^2 2x + \cos^2 x \cdot \cos^2 2x \\ &= (\sin^2 x + \cos^2 x)(\sin^2 2x + \cos^2 2x) - \sin^2 x \cdot \cos^2 2x - \sin^2 2x \cdot \cos^2 x \\ &= 1 - \sin^2 x \cdot \cos^2 2x - \sin^2 2x \cdot \cos^2 x \le 1. \end{align}$$

Let $\mathbf u(x) = (\sin 3x, \cos 3x) \in \mathbb R^2$. Then we have $|\mathbf u(x)|=1$. By Cauchy–Schwarz inequality:

$$\begin{align} \sin x \cdot \sin 2x \cdot \sin 3x + \cos x \cdot \cos 2x \cdot \cos 3x &= \mathbf u(x) \cdot \mathbf v(x) \\ &\le |\mathbf v(x)| \\ &\le 1,\end{align}$$

so for equality to hold, it in necessary (but not necessarily sufficient) that $|\mathbf v(x)| =1$, which means that both of these equations must hold: $$\begin{align} \sin x \cdot \cos 2x &= 0 \\ \sin 2x \cdot \cos x &= 0. \end{align}$$

By $\pi$-periodicity of the expression in $x$ (since $\sin (n(x+\pi)) = (-1)^n \sin (nx)$ and similarly for cosine), it's enough to find the solutions to this in the interval $x \in [0, \pi)$. Obviously, $x=0$ is one such solution. For all other $x$ in this interval, $\sin x \ne 0$ so we need $\cos 2x =0$ to satisfy the first equation. This has two solutions in this interval, namely $x \in \{ \frac{\pi}{4}, \frac{3\pi}{4} \}$, neither of which satisfy the second equation.

Since that was a necessary but not sufficient condition, you need to go back and check that $x=0$ does satisfy the equation (which is trivial). With that out of the way, the solutions are exactly $x = n \pi$ for integer $n$.

_{Note: This answer is mostly interesting because it proves a more general fact, namely $\sin \theta \sin \phi \sin \psi + \cos \theta \cos \phi \cos \psi \le 1$ for any angles $\theta,\phi,\psi$, and the method can be extended to longer products and gives necessary conditions to be checked for equality which are easy to work with. It's clearly more complicated than strictly required for this problem.}

Answer 4

$$\sin x \sin 2x \sin 3x + \cos x \cos 2x \cos 3x =1$$

We have $\sin x \sin 3x = \frac{\cos2x-\cos 4x}2$ and $\cos x \cos 3x = \frac{\cos 2x+\cos 4x}2$ from product-to-sum identities. Using this we get the original equation to the form

$$\sin2x (\cos2x-\cos4x)+\cos2x(\cos2x+\cos4x)=2 \tag{1}$$ We know that
$\sin 2x=2\frac{\sin x}{\cos x}\cos^2x = 2\frac{\tan x}{1+\tan^2x}$
$\cos 2x = 2\cos^2x-1 = \frac2{1+\tan^2x}-1=\frac{1-\tan^2x}{1+\tan^2x}$
$\cos 4x = \cos^4x - 6\sin^2x\cos^2x+\sin^4x = \frac{1-6\tan^2x+\tan^4x}{(1+\tan^2x)^2}$
(This is similar to Weierstrass substitution. I was also trying to use the substitution $2x=a$ in (1) and simplify the expression further, but I did not succeed using this path.)

With the substitution $t=\tan x$ we get $$\frac{2t}{1+t^2} \left( \frac{1-t^2}{1+t^2}-\frac{1-6t^2+t^4}{(1+t^2)^2}\right)+ \frac1{1+t^2} \left( \frac{1-t^2}{1+t^2}+\frac{1-6t^2+t^4}{(1+t^2)^2}\right)=2$$ $$2t((1-t^4)-(1-6t^2+t^4))+(1-t^4)+(1-6t^2+t^4)=2(1+t^2)^3$$ $$2t(6t^2-2t^4)+2-6t^2=2(1+3t^2+3t^4+t^6)$$ $$t(6t^2-2t^4)+1-3t^2=1+3t^2+3t^4+t^6$$ $$t^6+2t^5+3t^4-6t^3+6t^2=0$$ $$t^2(t^4+2t^3+3t^2-6t+6)=0$$

We could try to solve $t^4+2t^3+3t^2-6t+6=0$ by some of the methods for solving quartic equations. This is a lot of work to do by hand, but it can be done in principle.

Let us try t^4+2t^3+3t^2-6t+6 in WolframAlpha. WA says that there are no real roots for this equation.

Let us try whether we can show that this has no real roots without relying on a software.

$t^4+2t^3+3t^2-6t+6=t^4+2t^2+t^2+ 3(t^2-2t+1)-t^2=t^2(t+2)t+3(t-1)^2$ $\Rightarrow$ positive for $t>0$ and $t<-2$

For $t\in[-2,0]$ we have $0\le t^2\le 4$, hence $t^3\ge 4t$ and $t^4+2t^3+3t^2-6t+6\ge 3t^2-2t+6$. For the quadratic polynomial we have discriminant equal to $D=4-4\cdot 6\cdot 3 <0$, so this has no real roots and thus we get $$t^4+2t^3+3t^2-6t+6\ge 3t^2-2t+6 \ge 0 \qquad \text{for }t\in[0,2].$$

For $f(t)=t^4+2t^3+3t^2-6t+6$ we have $f'(t)=4t^3+6t^2+6t-6=4(t^3+\frac32t^2+\frac32t-\frac32)=4[(t+\frac12)^3]$.

So we have $t=0$ as the only solution, which yields $\tan x=0$, which gives $x=k\pi$.

Answer 5

My solution:

I. $\sin x\sin2x\sin 3x<0=>$ $ 1 < 1 - \sin x\sin2x\sin 3x = \cos x\cos 2x\cos 3x\leq 1$ False!

II. $\sin x\sin2x\sin 3x = 0 => \cos x\cos 2x\cos 3x\ = 1 =>... x=n\pi, n$ integer

III. $\sin x\sin2x\sin 3x>0=>$ $ 1 = |\sin x\sin2x\sin 3x + \cos x\cos 2x\cos 3x|\leq$ $ |\sin x\sin2x\sin3x| + |\cos x\cos 2x\cos 3x|=$ $ |\sin x\sin2x||\sin3x| + |\cos x\cos 2x||\cos 3x|\leq$ $ \leq|\sin x\sin 2x| + |\cos x\cos 2x| =\pm\sin x\sin2x \pm \cos x\cos 2x$= $\pm\cos (x\pm2x)\leq 1$.

a) $\cos x =\pm1 =>\sin x=0$ False!

b) $\cos 3x =\pm1 =>\sin 3x=0$ False!

Answer 6

Here is the worked out solution.

$\sin(x)\sin(2x)\sin(3x)+\cos(x)\cos(2x)\cos(3x)=1$

$\sin(x)[2\sin(x)\cos(x)][\sin(2x+x)]+\cos(x)[\cos^2(x)-\sin^2(x)][\cos(2x+x)]=1$

$2\sin^2(x)\cos(x)[\sin(2x)\cos(x)+\cos(2x)\sin(x)]+[\cos^3(x)-\cos(x)\sin^2(x)][\cos(2x)\cos(x)-\sin(2x)\sin(x)=1$

Those are my first few steps. Just continue pulling it apart using sum and difference formulas then at the end bring it all together using $\sin^2(x)+\cos^2(x)=1$.