Đáp án:
a) Vô nghiệm
b) $x=\dfrac{\pi }{4}+k\pi \left( k\in \mathbb{Z} \right)$
c) $\left[\begin{array}{l}x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{array}\right.\,\,\,\left(k\in\mathbb{Z}\right)$
d) $x=\dfrac{\pi }{12}+\dfrac{k\pi }{3}\,\,\,\left( k\in \mathbb{Z} \right)$
Giải thích các bước giải:
a) $6{{\sin }^{2}}3x+\cos 12x=14$
$\Leftrightarrow 6\cdot \dfrac{1-\cos 6x}{2}+\left( 2{{\cos }^{2}}6x-1 \right)=14$
$\Leftrightarrow 3-3\cos 6x+2{{\cos }^{2}}6x-1-14=0$
$\Leftrightarrow 2{{\cos }^{2}}6x-3\cos 6x-12=0$
Vô nghiệm vì $\cos 6x\in \left[ -1;1 \right]$
b) $4{{\sin }^{4}}x+12{{\cos }^{2}}x=7$
$\Leftrightarrow 4{{\left( {{\sin }^{2}}x \right)}^{2}}+12{{\cos }^{2}}x=7$
$\Leftrightarrow 4{{\left( 1-{{\cos }^{2}}x \right)}^{2}}+12{{\cos }^{2}}x-7=0$
$\Leftrightarrow 4\left( {{\cos }^{4}}x-2{{\cos }^{2}}x+1 \right)+12{{\cos }^{2}}x-7=0$
$\Leftrightarrow 4{{\cos }^{4}}x+4{{\cos }^{2}}x-3=0$
$\Leftrightarrow {{\cos }^{2}}x=\dfrac{1}{2}$ (nhận) hoặc ${{\cos }^{2}}x=-\dfrac{3}{2}$ (loại)
$\Leftrightarrow \cos x=\pm \dfrac{\sqrt{2}}{2}$
$\Leftrightarrow x=\dfrac{\pi }{4}+k\pi \left( k\in \mathbb{Z} \right)$
c) $5\sin 3x+\cos 6x+2=0$
$\Leftrightarrow 5\sin 3x+\left( 1-2{{\sin }^{2}}3x \right)+2=0$
$\Leftrightarrow -2{{\sin }^{2}}3x+5\sin 3x+3=0$
$\Leftrightarrow \sin 3x=-\dfrac{1}{2}$ (nhận) hoặc $\sin 3x=3$ (loại)
$\Leftrightarrow\left[\begin{array}{l}3x=-\dfrac{\pi }{6}+k2\pi\\3x=\dfrac{7\pi }{6}+k2\pi\end{array}\right.\,\,\,\left(k\in\mathbb{Z}\right)$
$\Leftrightarrow\left[\begin{array}{l}x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{array}\right.\,\,\,\left(k\in\mathbb{Z}\right)$
d) Giống câu b, khác mỗi ẩn $3x$ và $x$
Kết quả là $x=\dfrac{\pi }{12}+\dfrac{k\pi }{3}\,\,\,\left( k\in \mathbb{Z} \right)$
