Trigonometric Equations: Very Difficult Problems with Solutions

Solution:
Domain: [tex]\sin x\ne 0[/tex]. Multiplying by [tex]\sin ^2 x[/tex]:
[tex]3\cos^2 x+8\sin x\cos x+3\sin ^2 x=0\iff 3+4\sin 2x=0\iff\sin 2x=-\frac{3}{4}[/tex]
Let [tex]\beta=\arcsin \frac{3}{4}[/tex]. For [tex]x\in\left[0, 2\pi\right][/tex] we have [tex]2x\in\left[0, 4\pi\right][/tex], and there [tex]\sin 2x=-\frac{3}{4}[/tex] has four solutions:
[tex]2x=\pi+\beta, \ 2\pi-\beta, \ 3\pi+\beta, \ 4\pi-\beta[/tex]
None of them gives [tex]\sin x=0[/tex]. Their sum is [tex]10\pi[/tex], so the sum of the roots [tex]x[/tex] is [tex]5\pi[/tex].

Solution:
[tex]\cos A=\cos B[/tex] means [tex]A=\pm B+2k\pi[/tex]. With the plus sign:
[tex]x^2-x+1=x-1+2k\iff x^2-2x+2-2k=0\iff x=1\pm\sqrt{2k-1}[/tex]
([tex]k\ge 1[/tex]). A negative root needs [tex]\sqrt{2k-1}\gt 1[/tex], i.e. [tex]k\ge 2[/tex]; the largest such root is [tex]1-\sqrt{3}\approx-0. 73[/tex] (for [tex]k=2[/tex]). With the minus sign:
[tex]x^2-x+1=-x+1+2k\iff x^2=2k[/tex]
and the largest negative root is [tex]-\sqrt{2}\approx-1. 41[/tex]. Since [tex]1-\sqrt{3}\gt-\sqrt{2}[/tex], the answer is [tex]x=1-\sqrt{3}[/tex].

Solution:
Domain: [tex]\tan \left(x-\frac{\pi}{4}\right)\le 0[/tex], i.e. [tex]-\frac{\pi}{2}+k\pi\lt x-\frac{\pi}{4}\le k\pi[/tex], so [tex]x\in\left(-\frac{\pi}{4}+k\pi, \frac{\pi}{4}+k\pi\right][/tex]. The first factor is zero for [tex]x=\frac{\pi}{4}+k\pi[/tex]. For the second factor:
[tex]\sin x+\sin 5x+2\sin 3x=2\sin 3x\cos 2x+2\sin 3x=2\sin 3x(1+\cos 2x)=4\sin 3x\cos^2 x[/tex]
[tex]\cos x=0[/tex] gives [tex]x=\frac{\pi}{2}+k\pi[/tex], which is not in the domain. [tex]\sin 3x=0[/tex] gives [tex]x=\frac{k\pi}{3}[/tex]; since [tex]\frac{\pi}{3}\gt\frac{\pi}{4}[/tex], only [tex]x=k\pi[/tex] lies in the domain.

Solution:
Domain: [tex]\sin x\ne 0[/tex], [tex]\cos x\ne 0[/tex]. Multiplying by [tex]\sin x\cos x[/tex]:
[tex]8\sin x\cos^2 x=\sqrt{3}\cos x+\sin x[/tex]
The left side is [tex]4\sin 2x\cos x=2(\sin 3x+\sin x)[/tex]. So
[tex]2\sin 3x=\sqrt{3}\cos x-\sin x=2\sin \left(\frac{\pi}{3}-x\right)[/tex]
From [tex]\sin 3x=\sin \left(\frac{\pi}{3}-x\right)[/tex]: [tex]3x=\frac{\pi}{3}-x+2k\pi[/tex], so [tex]x=\frac{\pi}{12}+\frac{k\pi}{2}[/tex]; or [tex]3x=\pi-\frac{\pi}{3}+x+2k\pi[/tex], so [tex]x=\frac{\pi}{3}+k\pi[/tex]. Both series satisfy the domain.

Solution:
Domain: [tex]\cos 2x\ne 0[/tex], [tex]\cos 3x\ne 0[/tex]. Multiplying by [tex]\cos^2 2x\cos 3x[/tex]:
[tex]\sin ^2 2x\cos 3x+2\sin 2x\cos 2x\sin 3x-\cos^2 2x\cos 3x=0\iff\sin 4x\sin 3x-\cos 4x\cos 3x=0\iff\cos 7x=0[/tex]
Then [tex]x=\frac{(2k+1)\pi}{14}[/tex], and in [tex]\left[0, \pi\right][/tex] we get [tex]k=0, 1, \dots, 6[/tex] (seven values). For [tex]x=\frac{7\pi}{14}=\frac{\pi}{2}[/tex] we have [tex]\cos 3x=0[/tex], so it is rejected; for all the values [tex]\cos 2x\ne 0[/tex]. So there are [tex]6[/tex] roots.

Solution:
Let [tex]t=\sin x+\cos x=\sqrt{2}\sin \left(x+\frac{\pi}{4}\right)[/tex], [tex]|t|\le\sqrt{2}[/tex]. Then [tex]\sin x\cos x=\frac{t^2-1}{2}[/tex] and
[tex]\sin ^3 x+\cos^3 x=t(1-\sin x\cos x)=\frac{t\left(3-t^2\right)}{2}[/tex]
The equation becomes
[tex]2+3t-t^3=3t^2-3\iff t^3+3t^2-3t-5=0\iff(t+1)\left(t^2+2t-5\right)=0[/tex]
The roots [tex]t=-1\pm\sqrt{6}[/tex] are rejected, because [tex]\sqrt{6}-1\approx 1. 45\gt\sqrt{2}[/tex] and [tex]-1-\sqrt{6}\lt-\sqrt{2}[/tex]. So [tex]\sin \left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}[/tex]: [tex]x+\frac{\pi}{4}=-\frac{\pi}{4}+2k\pi[/tex] or [tex]x+\frac{\pi}{4}=\frac{5\pi}{4}+2k\pi[/tex], i.e. [tex]x=-\frac{\pi}{2}+2k\pi[/tex] or [tex]x=\pi+2k\pi[/tex].

Solution:
Domain: [tex]\sin x\cos x\ne 0[/tex]. The left side is [tex]2\sin \left(7x+\frac{\pi}{4}\right)\in\left[-2, 2\right][/tex]. The right side is [tex]t+\frac{1}{t}[/tex] with [tex]t=\tan x[/tex], so [tex]|\tan x+\cot x|\ge 2[/tex], with equality only for [tex]\tan x=\pm 1[/tex]. Case [tex]\tan x=1[/tex]: [tex]x=\frac{\pi}{4}+k\pi[/tex], the right side is [tex]2[/tex], but [tex]7x+\frac{\pi}{4}=2\pi+7k\pi[/tex] and the left side is [tex]0[/tex]; no roots. Case [tex]\tan x=-1[/tex]: [tex]x=-\frac{\pi}{4}+k\pi[/tex], the right side is [tex]-2[/tex], and we need
[tex]\sin \left(7x+\frac{\pi}{4}\right)=\sin \left(-\frac{3\pi}{2}+7k\pi\right)=-1[/tex]
This holds exactly for odd [tex]k=2m+1[/tex], so [tex]x=-\frac{\pi}{4}+(2m+1)\pi=\frac{3\pi}{4}+2m\pi[/tex].

Solution:
Domain: [tex]x\ge 0[/tex]. The left side is at least [tex]1[/tex] and the right side is at most [tex]1[/tex], so both are equal to [tex]1[/tex]: [tex]\cos x=1[/tex] and [tex]\sin \sqrt{2x}=0[/tex]. Then [tex]x=2k\pi[/tex] with an integer [tex]k\ge 0[/tex], and [tex]\sqrt{2x}=n\pi[/tex], i.e. [tex]x=\frac{n^2\pi^2}{2}[/tex] with an integer [tex]n\ge 0[/tex]. So [tex]2k\pi=\frac{n^2\pi^2}{2}[/tex], i.e. [tex]4k=n^2\pi[/tex]. If [tex]n\ne 0[/tex], then [tex]\pi=\frac{4k}{n^2}[/tex] would be rational, which is false. Hence [tex]n=k=0[/tex], and the only root is [tex]x=0[/tex].

Solution:
The left side is [tex]\ge 0[/tex] and the right side is [tex]\le 0[/tex], so both must be [tex]0[/tex]:
[tex]\cos 6x=1, \quad\sin (x-a)=-\frac{\sqrt{3}}{2}[/tex]
The first condition gives [tex]x=\frac{k\pi}{3}[/tex]. The second gives [tex]x-a=-\frac{\pi}{3}+2n\pi[/tex] or [tex]x-a=-\frac{2\pi}{3}+2n\pi[/tex]. Hence [tex]a=\frac{(k+1)\pi}{3}-2n\pi[/tex] or [tex]a=\frac{(k+2)\pi}{3}-2n\pi[/tex]: [tex]a[/tex] must be an integer multiple of [tex]\frac{\pi}{3}[/tex], and every such multiple works. In [tex]\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex] these are [tex]-\frac{\pi}{3}, 0, \frac{\pi}{3}[/tex].

Solution:
Since [tex]\sin 2x+\cos 2x-1=2\sin x\cos x-2\sin ^2 x[/tex], the equation is
[tex]\sin x(2\cos x-2\sin x-a)=0[/tex]
[tex]\sin x=0[/tex] gives the roots [tex]0[/tex] and [tex]\pi[/tex]. We need exactly one more root of [tex]g(x)=\cos x-\sin x=\frac{a}{2}[/tex], different from [tex]0[/tex] and [tex]\pi[/tex]. Since [tex]g(x)=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)[/tex], on [tex]\left[0, \frac{3\pi}{4}\right][/tex] it decreases from [tex]1[/tex] to [tex]-\sqrt{2}[/tex], and on [tex]\left[\frac{3\pi}{4}, \pi\right][/tex] it increases from [tex]-\sqrt{2}[/tex] to [tex]-1[/tex].
For [tex]-1\lt\frac{a}{2}\lt 1[/tex] there is one new root. For [tex]\frac{a}{2}=-1[/tex] the roots of [tex]g[/tex] are [tex]\frac{\pi}{2}[/tex] and [tex]\pi[/tex], so again one new root. For [tex]\frac{a}{2}=-\sqrt{2}[/tex] the only new root is [tex]\frac{3\pi}{4}[/tex]. For [tex]-\sqrt{2}\lt\frac{a}{2}\lt-1[/tex] there are two new roots, for [tex]\frac{a}{2}=1[/tex] only [tex]x=0[/tex], and otherwise none. So [tex]a\in\left\{-2\sqrt{2}\right\}\cup\left[-2, 2\right)[/tex].

Solution:
Let [tex]S[/tex] be the left side ([tex]1010[/tex] terms). If [tex]\sin x=0[/tex], all terms are equal to [tex]1[/tex] or all are equal to [tex]-1[/tex], so [tex]S=\pm 1010\ne 0[/tex]. If [tex]\sin x\ne 0[/tex], multiply by [tex]2\sin x[/tex] and use
[tex]2\sin x\cos(2j-1)x=\sin 2jx-\sin (2j-2)x[/tex]
The sum telescopes:
[tex]2\sin x\times S=\sin 2020x[/tex]
So [tex]S=0[/tex] exactly when [tex]\sin 2020x=0[/tex] and [tex]\sin x\ne 0[/tex], i.e. [tex]x=\frac{k\pi}{2020}[/tex], where [tex]k[/tex] is not divisible by [tex]2020[/tex].

Solution:
In this interval [tex]\cos x\ne 0[/tex]. Since
[tex]\tan 3x-\tan x=\frac{\sin 2x}{\cos 3x\cos x}=\frac{2\sin x}{\cos 3x}[/tex]
we have [tex]\frac{\sin x}{\cos 3x}=\frac{1}{2}(\tan 3x-\tan x)[/tex]. In the same way [tex]\frac{\sin 3x}{\cos 9x}=\frac{1}{2}(\tan 9x-\tan 3x)[/tex] and [tex]\frac{\sin 9x}{\cos 27x}=\frac{1}{2}(\tan 27x-\tan 9x)[/tex]. The sum is
[tex]\frac{1}{2}(\tan 27x-\tan x)=\frac{\sin 26x}{2\cos x\cos 27x}[/tex]
So [tex]\sin 26x=0[/tex], [tex]x=\frac{k\pi}{26}[/tex], and the interval gives [tex]k=-12, \dots, 12[/tex]. For these [tex]k[/tex] the denominators [tex]\cos 3x[/tex], [tex]\cos 9x[/tex], [tex]\cos 27x[/tex] are not zero: [tex]\cos 3^m x=0[/tex] would need [tex]3^m k=13(2j+1)[/tex], which is impossible for [tex]|k|\lt 13[/tex]. So all [tex]25[/tex] values are roots.