d) ĐK: $\sin x \neq 0, \cos x \neq 0$ hay $x \neq k\pi$ và $x \neq \pi/2 + k\pi$.
Ptrinh tương đương vs
$$2\tan x - \dfrac{3}{\tan x} -2 = 0$$
$$<-> 2\tan^2x -2\tan x -3 = 0$$
<-> $\tan x = \dfrac{1 \pm \sqrt{7}}{2}$
Vậy $x = arctan(\dfrac{1 \pm \sqrt{7}}{2})$.
e) Ptrinh tương đương vs
$\cos^2x - \dfrac{3}{2} \sin (2x) + 3 = 1$
$<->\dfrac{1 + 2\cos(2x)}{2} - \dfrac{3}{2} \sin (2x) = -2$
$<-> 2\cos(2x) - 3\sin(2x) = -5$
Chia cả 2 vế cho $\sqrt{2^2 + 3^2} = \sqrt{13}$
$\dfrac{2}{\sqrt{13}}\cos(2x) - \dfrac{3}{\sqrt{13}} \sin(2x) = -\dfrac{5}{\sqrt{13}}$
Đặt $\sin a = \dfrac{2}{\sqrt{13}}$, $\cos a = \dfrac{3}{\sqrt{13}}$, $\sin b = -\dfrac{5}{\sqrt{13}}.
Khi đó, ta có
$\sin(a-2x) = \sin b$
<->$ a-2x = b + 2k\pi$ hoặc $a - 2x = \pi -b + 2k\pi$
<->$ x = \dfrac{a-b}{2} + k\pi$ hoặc $x = \dfrac{a+b-\pi}{2} + k\pi$.
c) Ptrinh tương đương vs
$-\dfrac{1}{4} + \sin^2x = (1-\sin^2x)^2$
<->$ -\dfrac{1}{4} + \sin^2x = 1 + \sin^4x -2\sin^2x$
<->$ \sin^4x -3\sin^2x +\dfrac{5}{4} = 0$
<->$ \sin^2x = \dfrac{1}{2}$ hoặc $\sin^2x = \dfrac{5}{2}$
<->$ \sin x = \dfrac{1}{\sqrt{2}}$ hoặc $\sin x = \sqrt{\dfrac{5}{2}}$ (loại do $-1 \leq \sin x \leq 1$.
Vậy $x = \dfrac{\pi}{4} + 2k\pi$ hoặc $x = \dfrac{3\pi}{4} + 2k\pi$.
