Trigonometric functions - finding solutions

Question

Question:

Find the general solution for the equation: $$\sin x + 2\sin2x - \sin3x = 3$$

Approach: Well using the identity of $\sin a - \sin b $, I merged together $\sin x - \sin3x$ And as $\sin 2x = 2\sin x\cos x$, I got up till here: $$2\sin x(2\cos x - \cos 2x) = 3$$

Edit:

Considering $\sin x - \sin 3x = -2\cos 2x\sin x$ The equation becomes: $$-2\cos 2x\sin x+2\sin 2x = 3$$ $$(-2\sin x)\cos 2x +2\sin 2x = 3$$ The maximum value of the equation $asinx + bcosx$ is $\sqrt{a^2 + b^2}$ Using this in here: (considering $b = -2\sin x$ and $a = 2$) $$3 \leq \sqrt{4\sin^2 x + 4}$$ $$3/2 \leq \sqrt{1 + sin^2x}$$ Obviously cannot happen as maximum value of $\sqrt{1 + sin^2x}$ is $\sqrt2$

Use $sinx=\sqrt{1-cos^2x}$, isolate the square-root-term and square the equality. — Peter, Jan 21 '15 at 16:04
@Peter I do believe there to be an easier way to do this. This method will make the question way too long. — Gummy bears, Jan 21 '15 at 16:08
@Peter I just found an easier method. Simplifies the question much more. — Gummy bears, Jan 21 '15 at 16:17
I do not understand the last step in the edit. But anyway, the right side cannot exceed $\sqrt{8}$, so the idea seems to work. — Peter, Jan 21 '15 at 16:32
@Peter Well take 4 common, leaving $1+\sin^2x$ which is equal to $\cos^2x$ Thus we have $\sqrt{4cos^2x}$ — Gummy bears, Jan 21 '15 at 16:34
@Peter Oh true... Well it's still valid. maximum value of $\sqrt{\sin^2 x + 1}$ is $\sqrt{2}$ which is still less than 1.5 — Gummy bears, Jan 21 '15 at 16:38
Look at my comment, I got the contradiction in a similar way. — Peter, Jan 21 '15 at 16:39
@Peter Squaring it would have proven much more difficult as we would have to square a term in the form $(a + b + c)$ — Gummy bears, Jan 21 '15 at 16:41

score 3 · Answer 1 · answered Jan 21 '15 at 16:28

3

By setting $z=e^{ix}\in S^1$, the given equation can be written as: $$ \Im\left(z+2z^2-z^3\right) = 3 \tag{1}$$ that implies: $$ \left\|z+2z^2-z^3\right\| = \left\| 2+\frac{1}{z}-z \right\|\geq 3\tag{2}$$ but that cannot happen since: $$ \left\| 2+\frac{1}{z}-z \right\| = \left\| 2-2i\sin x\right\|\leq 2\sqrt{2} < 3,\tag{3} $$ so there are no real solutions.

answered Jan 21 '15 at 16:28

Jack D'Aurizio

353,855

Sorry but in the first step, what is that thingy in the starting? Looks like a 'J'. – Gummy bears Jan 21 '15 at 16:30
@Gummybears: it stands for imaginary part. Have you studied complex numbers? – Jack D'Aurizio Jan 21 '15 at 16:31
By the way, this proof just uses the Cauchy-Schwarz inequality like yours. – Jack D'Aurizio Jan 21 '15 at 16:32
Yes I have studied complex numbers. Just never studied that symbol. Always used Im(z) to represent the imaginary part. – Gummy bears Jan 21 '15 at 16:33
Ok, just a notation matter, $\Im(z)$ stands for $\operatorname{Im}(z)$. – Jack D'Aurizio Jan 21 '15 at 16:34
Hmmmm... Where did the inequality appear from? Lost you there in the second step. And also, you took $z^2$ as a common factor, where did that factor disappear? – Gummy bears Jan 21 '15 at 16:37
@Gummybears: everything follows from $\sin x = \Im(e^{ix})$, so $\sin(2x)=\Im(z^2)$ and so on. – Jack D'Aurizio Jan 21 '15 at 16:47
@Gummybears: moreover, $z$ is a complex number with modulus one, so $|z^2|=1$ can be factored away. – Jack D'Aurizio Jan 21 '15 at 16:48
What's $S^1$ stand for? – GFauxPas Jan 21 '15 at 17:14
@GFauxPas: the unit circle. – Jack D'Aurizio Jan 21 '15 at 17:16
Ohhhh.... the the fact that the modulus is 1 clears it up. – Gummy bears Jan 22 '15 at 11:58
In the last step, where did $2 - 2\sin x$ appear from? – Gummy bears Jan 22 '15 at 12:05
@Gummybears: if $z=e^{ix}$, then $\frac{1}{z}-z = 2i\sin x$. – Jack D'Aurizio Jan 22 '15 at 12:07
1

Ohhh Okay. In the end it comes out to the same inequality. Really shows how closely complex numbers are trigonometry are related. – Gummy bears Jan 22 '15 at 12:22

Alex Silva · Answer 2 · 2015-01-21T18:10:25.197

By using the trigonometric identities:

$$\sin(2x) = 2\sin(x)\cos(x)$$ $$\sin(3x) = 3\sin(x)-4\sin^{3}(x),$$ we have

$$\sin(x) +4\sin(x)\cos(x) -3\sin(x) + 4\sin^3(x) = 3 \implies \cos(x) = \frac{3 + 2\sin(x) -4\sin^3(x)}{4\sin(x)}.$$

By squaring the equation, the solutions are kept. Hence,

$$1-\sin^2(x) = \frac{16\sin^6(x)-16\sin^4(x)-24\sin^3(x)+4\sin^2(x)+12\sin(x) +9}{16\sin^2(x)}.$$

Take $y = \sin(x)$. Thus,

$$(1-y^2) = \frac{16y^6-16y^4-24y^3+4y^2 +12y+9}{16y^2}$$ $$\implies 16y^6-24y^3-12y^2 +12y+9 =0,$$ which has no real solutions.

score 0 · Answer 3 · answered Jan 21 '15 at 16:04

0

There are no real solutions. $s = \sin(x)$ must satisfy $16s^6-24s^3-12s^2+12s+9=0$, which has no real root.

answered Jan 21 '15 at 16:04

Robert Israel

448,999

How did you reach that conclusion though? – Gummy bears Jan 21 '15 at 16:07
I used Maple's sturm procedure to count the real roots. – Robert Israel Jan 21 '15 at 23:16

Trigonometric functions - finding solutions

3 Answers3