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Find general solution of the given trigonometric equation:

$\sin^24x + \cos^2x = 2 \sin4x \cos^2x$

I tried converting the whole equation in the form of $2x$ and got a pretty complicated equation involving $\sin2x, \cos2x, \sin^22x$ and $\cos^22x$. I have got no idea how to further solve that equation.

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3 Answers 3

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Rearrange the equation as follows,

$$\sin^24x + \cos^2x - 2 \sin4x \cos^2x$$ $$=(\cos^2x - \sin 4x)^2 -\cos^4 x+\cos^2x$$ $$=(\cos^2x - \sin 4x)^2 +\cos^2x \sin^2x$$ $$=(\cos^2x - 2\sin 2x\cos2x)^2 +\frac14\sin^2 2x=0$$

where both terms vanish, leading to the following system of equations, $$\sin2x =0$$ $$\cos^2x=2\cos2x\sin2x=0$$

Thus, the valid solutions are $x=\frac\pi2+n\pi$.

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  • $\begingroup$ $(\cos^2x - \sin 4x)^2 -\cos^4 x+\cos^2x$ How did you get this? $\endgroup$ Feb 24, 2020 at 19:05
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    $\begingroup$ @Aditya Jain - Note $(\cos^2x - \sin 4x)^2 = \cos^4x -2\sin4x\cos^2+\sin^24x$. Then the term $\cos^4x$ cancels. $\endgroup$
    – Quanto
    Feb 24, 2020 at 19:17
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Let $x$ be a solution of the given equation. Associate the equation of degree two in $T$, $$ \tag{$*$} T^2 -2T\cos^2 x+\cos^2 x=0\ . $$ It has the solution $T_1=\sin 4x$, so the equation $(*)$ has discriminant $\ge 0$ (w.r.t. $T$), we obtain then $\cos^4 x-\cos^2 x\ge 0$. This is equivalent to $-\sin^2 x\cos^2x\ge 0$. So either $\sin x$ or $\cos x$ vanishes.

  • The case $\sin x=0$: We obtain $\sin 4x=0$ and $\cos x=\pm 1$, no solutions.

  • The case $\cos x=0$: We obtain also $\sin 4x=2\sin 2x\cos 2x=4\sin x\cos x\cos 2x= 0$. This case always delivers solutions. So $x$ is an odd multiple of $\pi/2$.


Later edit: After the comment, here is an alternative way to finish by remaining inside trigonometry: $$ \begin{aligned} &\sin^2 4x + \cos^2x - 2 \sin4x \cos^2x \\ &\qquad= 4\sin^2 2x\cos^22x + \cos^2x - 4 \sin2x\cos 2x \cos^2x \\ &\qquad= 16\sin^2 x\cos^2 x\cos^22x + \cos^2x - 8 \sin x\cos x\cos 2x \cos^2x \\ &\qquad= \cos^2 x\Big(\ 16\sin^2 x\cos^22x - 8 \sin x\cos x\cos 2x + 1 \ \Big) \\ &\qquad= \cos^2 x\Big(\ 16\sin^2 x\cos^22x - 8 \sin x\cos x\cos 2x + \cos^2x\ +\sin^2x \ \Big) \\ &\qquad= \cos^2 x \underbrace {\Big(\ (4\sin x\cos 2x - \cos x)^2 +\sin^2x \ \Big)}_{\ge 0} \ . \end{aligned} $$ The first / last expression in the chain is zero, iff

  • either $\cos x=0$,
  • or the parenthesis above vanishes, for this, we need in particular $\sin x =0$, which implies $\sin 4x=0$ and $\cos x=\pm 1$, so we never get a solution.

(This is in essence the same solution...)

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  • $\begingroup$ Thanks. That is a great approach to the question. But can this be solved using just trigonometric identities as I am dealing with a chapter on trigonometric equations and identities? $\endgroup$ Feb 24, 2020 at 18:19
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$\sin^24x + \cos^2x = 2 \sin4x \cos^2x$

$\sin^24x-2 \sin4x \cos^2x + \cos^2x = 0$

$\sin^24x-2 \sin4x \cos^2x +\cos^4x-\cos^4x+ \cos^2x = 0$

$(\sin4x-\cos^2x)^2 +\cos^2x(1-\cos^2x) = 0$

$(\sin4x-\cos^2x)^2 \geqslant 0$

$\cos^2x(1-\cos^2x) \geqslant 0$

$\Rightarrow \sin4x-\cos^2x=\cos^2x(1-\cos^2x) = 0$

$\ cos^2x = 1 \Rightarrow\sin4x \neq 1 \; \varnothing$

$\ cosx = 0 \Rightarrow\sin4x= 0 \;OK$

$ x =\frac {\pi}{2}+\pi*n\in \ {Z} $

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