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I am stuck on a trigonometric problem which is as follow:

Solve $\sin^2 x+ \sin^2 2x-\sin^2 3x-\sin^2 4x=0$ for $x$.

these kind of problems tease me always. Please help me. To get a rough idea of how can I approach such problems.

super saiyan
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  • Can't you use trig. identity to write everything in $\sin(x)$? –  Jan 02 '17 at 11:32
  • Hint: $\sin 2x = 2 \sin x \cos x$ and $\sin (a+b) = \sin a \cos b + \sin b \cos a$. – YukiJ Jan 02 '17 at 11:32
  • @YukiJ, I tried that, but it will make a 4th degree equation in terms of sin... – super saiyan Jan 02 '17 at 11:36
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    Maybe helpful

    $$\sin^2 x+ \sin^2 2x-\sin^2 3x-\sin^2 4x=$$

    $$\sin^2 x-\sin^2 3x+ \sin^2 2x-\sin^2 4x=$$

    $$(\sin x-\sin 3x)(\sin x+\sin 3x)+ (\sin 2x-\sin 4x)(\sin 2x+\sin 4x)=...$$

     – Amin235 Jan 02 '17 at 11:50

2 Answers2

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First linearise with: $$\sin^2 u=\frac{1-\cos 2u}2.$$ Simplifying yields the equation $$\cos 2x+\cos 4x=\cos6x+\cos 8x.$$ Then the factorisation formula: $$\cos p+\cos q=2\cos\frac{p+q}2\cos\frac{p-q}2 $$ lets you rewrite the equation, after some simplification, as $$\cos x\cos 3x=\cos x\cos 7x\iff \begin{cases}\cos x=0\quad\text{or}\\\cos 3x=\cos 7x.\end{cases}$$ Solutions to the first equation: $\qquad x\equiv \dfrac\pi2\mod\pi$.

Solutions to the second equation: $$7x\equiv \pm 3x\mod 2\pi\iff\begin{cases} 4x\equiv 0\mod 2\pi\\10x\equiv 0\mod2\pi\end{cases}\iff\begin{cases} x\equiv 0\mod \dfrac\pi2\\[1ex]x\equiv 0\mod\dfrac\pi5\end{cases} $$ The solutions in the second series are redundant either with the first or the third series. The set of solutions can ultimately be described as: $$x\equiv \dfrac\pi2\mod\pi,\quad x\equiv 0\mod\dfrac\pi5. $$

Bernard
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HINT:

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin^23x-\sin^2x+\sin^24x-\sin^22x=\sin2x(\sin4x+\sin6x)$$

Now use Prosthaphaeresis Formulas $\sin4x+\sin6x=2\sin5x\cos x$

Community
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lab bhattacharjee
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