# Find general solution of the given trigonometric equation

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.

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 $$x=\frac\pi2+n\pi$$.

• $(\cos^2x - \sin 4x)^2 -\cos^4 x+\cos^2x$ How did you get this? Feb 24, 2020 at 19:05
• @Aditya Jain - Note $(\cos^2x - \sin 4x)^2 = \cos^4x -2\sin4x\cos^2+\sin^24x$. Then the term $\cos^4x$ cancels. Feb 24, 2020 at 19:17

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...)

• 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? Feb 24, 2020 at 18:19

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

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

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

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

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

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

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

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

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

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