Đáp án: $x=\pm \dfrac{\pi }{3}+k\pi \left( k\in \mathbb{Z} \right)$
Giải thích các bước giải:
Điều kiện: $\cos x\ne 0\Rightarrow x\ne \dfrac{\pi }{2}+k\pi \left( k\in \mathbb{Z} \right)$
${{\sin }^{2}}x+{{\sin }^{2}}x.{{\tan }^{2}}x=3$
$\Leftrightarrow {{\sin }^{2}}x+{{\sin }^{2}}x.\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}=3$
$\Leftrightarrow {{\sin }^{2}}x.{{\cos }^{2}}x+{{\sin }^{4}}x=3{{\cos }^{2}}x$
$\Leftrightarrow {{\sin }^{2}}x\left( 1-{{\sin }^{2}}x \right)+{{\sin }^{4}}x=3\left( 1-{{\sin }^{2}}x \right)$
$\Leftrightarrow {{\sin }^{2}}x-{{\sin }^{4}}x+{{\sin }^{4}}x=3-3{{\sin }^{2}}x$
$\Leftrightarrow 4{{\sin }^{2}}x=3$
$\Leftrightarrow 4\cdot \dfrac{1-\cos 2x}{2}=3$
$\Leftrightarrow 2-2\cos 2x=3$
$\Leftrightarrow 2\cos 2x=-1$
$\Leftrightarrow \cos 2x=-\dfrac{1}{2}$
$\Leftrightarrow 2x=\pm \dfrac{2\pi }{3}+k2\pi \left( k\in \mathbb{Z} \right)$
$\Leftrightarrow x=\pm \dfrac{\pi }{3}+k\pi \left( k\in \mathbb{Z} \right)$
