Trigonometric Equations: Difficult Problems with Solutions

Solution:
Domain: [tex]\cos x\ne 0[/tex] (then also [tex]\sin x\ne-1[/tex]). With a common denominator:
[tex]\tan x+\frac{\cos x}{1+\sin x}=\frac{\sin x(1+\sin x)+\cos^2 x}{\cos x(1+\sin x)}=\frac{1+\sin x}{\cos x(1+\sin x)}=\frac{1}{\cos x}[/tex]
So [tex]\frac{1}{\cos x}=2[/tex], [tex]\cos x=\frac{1}{2}[/tex] and [tex]x=\pm\frac{\pi}{3}+2k\pi[/tex]. These values belong to the domain.

Solution:
Write [tex]\sin x=\sin x\left(\sin ^2 x+\cos^2 x\right)[/tex]. The equation becomes homogeneous of degree [tex]3[/tex]:
[tex]3\cos^3 x-\sin ^2 x\cos x-\sin ^3 x-\sin x\cos^2 x=0[/tex]
If [tex]\cos x=0[/tex], then [tex]\sin ^3 x=0[/tex], which is impossible, so we divide by [tex]\cos^3 x[/tex]. With [tex]t=\tan x[/tex]:
[tex]t^3+t^2+t-3=0\iff(t-1)\left(t^2+2t+3\right)=0[/tex]
The quadratic factor has no real roots, so [tex]\tan x=1[/tex] and [tex]x=\frac{\pi}{4}+k\pi[/tex].

Solution:
Divide by [tex]2[/tex]:
[tex]\cos 2x=\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x=\cos\left(x-\frac{\pi}{6}\right)\iff 2x=\pm\left(x-\frac{\pi}{6}\right)+2k\pi[/tex]
With the plus sign: [tex]x=-\frac{\pi}{6}+2k\pi[/tex]. With the minus sign: [tex]3x=\frac{\pi}{6}+2k\pi[/tex], so [tex]x=\frac{\pi}{18}+\frac{2k\pi}{3}[/tex].

Solution:
Let [tex]t=\cos x-\sin x[/tex], [tex]|t|\le\sqrt{2}[/tex]. Then [tex]t^2=1-\sin 2x[/tex], so [tex]\sin 2x=1-t^2[/tex] and the equation becomes
[tex]t^2-6t+5=0[/tex]
with roots [tex]t=1[/tex] and [tex]t=5[/tex]; the root [tex]t=5[/tex] is impossible. From
[tex]\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)=1\iff x+\frac{\pi}{4}=\pm\frac{\pi}{4}+2k\pi[/tex]
we get [tex]x=2k\pi[/tex] or [tex]x=-\frac{\pi}{2}+2k\pi[/tex].

Solution:
Let [tex]t=\cos 2x[/tex]. Then [tex]\sin ^2 x=\frac{1-t}{2}[/tex], so [tex]16\sin ^6 x=2(1-t)^3[/tex], and [tex]\cos 4x=2t^2-1[/tex]. The left side is
[tex]2\left(1-3t+3t^2-t^3\right)+6t-6t^2+3=5-2t^3[/tex]
So [tex]5-2t^3=\frac{19}{4}[/tex], [tex]t^3=\frac{1}{8}[/tex] and [tex]t=\frac{1}{2}[/tex]. From [tex]\cos 2x=\frac{1}{2}[/tex]: [tex]2x=\pm\frac{\pi}{3}+2k\pi[/tex], [tex]x=\pm\frac{\pi}{6}+k\pi[/tex].

Solution:
Use [tex]\cos 3x=4\cos^3 x-3\cos x[/tex] and [tex]\sin 3x=3\sin x-4\sin ^3 x[/tex]. The terms [tex]\pm 4\sin ^3 x\cos^3 x[/tex] cancel, and the left side becomes
[tex]3\sin x\cos^3 x-3\sin ^3 x\cos x=3\sin x\cos x\left(\cos^2 x-\sin ^2 x\right)=\frac{3}{2}\sin 2x\cos 2x=\frac{3}{4}\sin 4x[/tex]
So [tex]\sin 4x=1[/tex], [tex]4x=\frac{\pi}{2}+2k\pi[/tex] and [tex]x=\frac{\pi}{8}+\frac{k\pi}{2}[/tex].

Solution:
Write [tex]\sin x\sin 3x=\frac{1}{2}(\cos 2x-\cos 4x)[/tex] and [tex]\sin 4x=2\sin 2x\cos 2x[/tex]. The equation becomes
[tex]\frac{1}{2}\sin 2x(\cos 2x-\cos 4x)=\frac{1}{2}\sin 2x\cos 2x\iff\sin 2x\cos 4x=0[/tex]
[tex]\sin 2x=0[/tex] gives [tex]x=\frac{k\pi}{2}[/tex]; [tex]\cos 4x=0[/tex] gives [tex]4x=\frac{\pi}{2}+k\pi[/tex], [tex]x=\frac{\pi}{8}+\frac{k\pi}{4}[/tex]. Do not divide by [tex]\sin 2x[/tex]: this loses roots.

Solution:
The left side is non-negative, so we need [tex]\cos x\le 0[/tex]. Squaring:
[tex]1+\sin x=\cos^2 x=1-\sin ^2 x\iff\sin x(\sin x+1)=0[/tex]
If [tex]\sin x=0[/tex], then [tex]\cos x=\pm 1[/tex], and the condition [tex]\cos x\le 0[/tex] leaves only [tex]\cos x=-1[/tex]: [tex]x=\pi+2k\pi[/tex]. If [tex]\sin x=-1[/tex], then [tex]\cos x=0[/tex], which is allowed: [tex]x=-\frac{\pi}{2}+2k\pi[/tex]. The values [tex]x=2k\pi[/tex] are extraneous.

Solution:
Write [tex]3\cos^2 x-2=1-3\sin ^2 x[/tex].
Case [tex]0\le x\le\frac{\pi}{2}[/tex]: [tex]\sin x\ge 0[/tex], so [tex]\sin ^2 x=1-3\sin ^2 x[/tex], [tex]\sin ^2 x=\frac{1}{4}[/tex], [tex]\sin x=\frac{1}{2}[/tex] and [tex]x=\frac{\pi}{6}[/tex]. Case [tex]-\frac{\pi}{2}\le x\lt 0[/tex]: [tex]\sin x\lt 0[/tex], so [tex]-\sin ^2 x=1-3\sin ^2 x[/tex], [tex]\sin ^2 x=\frac{1}{2}[/tex], [tex]\sin x=-\frac{\sqrt{2}}{2}[/tex] and [tex]x=-\frac{\pi}{4}[/tex].

Solution:
Group the terms:
[tex]2\sin x(\cos x+1)+(\cos x+1)=(\cos x+1)(2\sin x+1)=0[/tex]
[tex]\cos x=-1[/tex] gives [tex]x=\pi+2k\pi[/tex]; in [tex]\left(0, 5\right)[/tex] only [tex]x=\pi\approx 3. 14[/tex]. [tex]\sin x=-\frac{1}{2}[/tex] gives [tex]x=\frac{7\pi}{6}+2k\pi[/tex] or [tex]x=\frac{11\pi}{6}+2k\pi[/tex]; in [tex]\left(0, 5\right)[/tex] only [tex]\frac{7\pi}{6}\approx 3. 67[/tex], because [tex]\frac{11\pi}{6}\approx 5. 76\gt 5[/tex]. So there are [tex]2[/tex] roots.

Solution:
Since [tex]2\cos^2 3x-1=\cos 6x[/tex], the equation is
[tex]\cos 6x-\cos 4x=0\iff-2\sin 5x\sin x=0[/tex]
[tex]\sin x=0[/tex] gives [tex]x=k\pi[/tex] and [tex]\sin 5x=0[/tex] gives [tex]x=\frac{k\pi}{5}[/tex]; the first series is part of the second. In [tex]\left[0, \frac{\pi}{2}\right][/tex] the roots are [tex]0, \frac{\pi}{5}, \frac{2\pi}{5}[/tex] (the next one, [tex]\frac{3\pi}{5}[/tex], is too big). So there are [tex]3[/tex] roots.

Solution:
Since [tex]-1\le\cos 4x\le 1[/tex], the left side satisfies [tex]5\le 8-3\cos 4x\le 11[/tex]. Since [tex]1\le 3-2\sin x\le 5[/tex], the right side satisfies [tex]1\le\frac{5}{3-2\sin x}\le 5[/tex]. Equality is possible only when both sides equal [tex]5[/tex]:
[tex]\cos 4x=1, \quad 3-2\sin x=1\iff\sin x=1[/tex]
Then [tex]x=\frac{\pi}{2}+2k\pi[/tex], and for these values [tex]4x=2\pi+8k\pi[/tex], so [tex]\cos 4x=1[/tex] holds.

Solution:
Domain: [tex]\cos x\ne 0[/tex], [tex]\cos 2x\ne 0[/tex], [tex]\cos 3x\ne 0[/tex]. Using [tex]\tan a+\tan b=\frac{\sin (a+b)}{\cos a\cos b}[/tex]:
[tex]\frac{\sin 3x}{\cos x\cos 2x}-\frac{\sin 3x}{\cos 3x}=\sin 3x\times \frac{\cos 3x-\cos x\cos 2x}{\cos x\cos 2x\cos 3x}=\frac{-\sin 3x\sin x\sin 2x}{\cos x\cos 2x\cos 3x}[/tex]
So [tex]\sin 3x\sin 2x\sin x=0[/tex]. [tex]\sin 3x=0[/tex] gives [tex]x=\frac{k\pi}{3}[/tex], and all these values are in the domain. [tex]\sin x=0[/tex] gives [tex]x=k\pi[/tex], already included. [tex]\sin 2x=0[/tex] adds [tex]x=\frac{\pi}{2}+k\pi[/tex], but there [tex]\cos x=0[/tex], so these values are rejected.

Solution:
Domain: [tex]\sin x\ne 0[/tex] (then also [tex]\cos x\ne 1[/tex]). Multiplying by [tex]\sin ^2 x(1-\cos x)[/tex]:
[tex]\cos^2 x-\cos^3 x=\sin ^2 x-\sin ^3 x[/tex]
[tex]\left(\cos^2 x-\sin ^2 x\right)-\left(\cos^3 x-\sin ^3 x\right)=(\cos x-\sin x)(\cos x+\sin x-1-\sin x\cos x)=-(\cos x-\sin x)(1-\sin x)(1-\cos x)=0[/tex]
[tex]\cos x=\sin x[/tex] gives [tex]x=\frac{\pi}{4}+k\pi[/tex]; [tex]\sin x=1[/tex] gives [tex]x=\frac{\pi}{2}+2k\pi[/tex]; [tex]\cos x=1[/tex] means [tex]\sin x=0[/tex] and is rejected.

Solution:
An equation [tex]p\cos x+q\sin x=r[/tex] with [tex]p^2+q^2\gt 0[/tex] has a solution exactly when [tex]r^2\le p^2+q^2[/tex] (divide by [tex]\sqrt{p^2+q^2}[/tex] and use an auxiliary angle). Here
[tex]p^2+q^2=(a-1)^2+(a+1)^2=2a^2+2\gt 0[/tex]
The condition is [tex]4a^2\le 2a^2+2[/tex], i.e. [tex]a^2\le 1[/tex] and [tex]-1\le a\le 1[/tex].

Solution:
Domain: [tex]\cos x\ne 0[/tex], [tex]\cos 2x\ne 0[/tex], [tex]\cos 3x\ne 0[/tex], [tex]\cos 4x\ne 0[/tex]. From
[tex]\tan 3x=\frac{\tan x+\tan 2x}{1-\tan x\tan 2x}[/tex]
(here [tex]1-\tan x\tan 2x=\frac{\cos 3x}{\cos x\cos 2x}\ne 0[/tex]) we get [tex]\tan x+\tan 2x+\tan x\tan 2x\tan 3x=\tan 3x[/tex]. So the equation becomes [tex]\tan 4x=0[/tex], i.e. [tex]x=\frac{k\pi}{4}[/tex]. For odd [tex]k[/tex] we have [tex]\cos 2x=0[/tex], and for [tex]x=\frac{m\pi}{2}[/tex] with odd [tex]m[/tex] we have [tex]\cos x=0[/tex]. Only [tex]x=k\pi[/tex] remain, and they satisfy all conditions.