Đáp án:
Giải thích các bước giải:
a) `2sin^2 x-sin\ x-1=0`
`⇔ 2sin^2 x-2sin\ x+sin\ x-1=0`
`⇔ (sin\ x-1)(2sin\ x+1)=0`
`⇔` \(\left[ \begin{array}{l}\sin\ x-1=0\\2\sin\ x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{7\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
f) `2cos^2 x+3cos\ x-2=0`
`⇔ 2cos^2 x+4cos\ x-cos\ x-2=0`
`⇔ (2cos\ x-1)(cos\ x+2)=0`
`⇔` \(\left[ \begin{array}{l}2\cos\ x-1=0\\\cos\ x+2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\cos\ x=\dfrac{1}{2}\\\cos\ x=-2\ (\text{Loại vì cos x $\in$ [-1;1]})\end{array} \right.\)
`⇔ x=\pm \frac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})`
h) `tan^2\ x+2\sqrt{3}tan\ x+3=0`
`⇔ (tan\ x)^2+2\sqrt{3}tan\ x+(\sqrt{3})^2=0`
`⇔ (tan\ x+\sqrt{3})^2=0`
`⇔ tan\ x=-\sqrt{3}`
`⇔ x=-\frac{\pi}{3}+k\pi\ (k \in \mathbb{Z})`
i) `3cot^2 x+2\sqrt{3}cot\ x+1=0`
`⇔ (\sqrt{3}cot\ x)^2+2\sqrt{3}cot\ x+1=0`
`⇔ (\sqrt{3}cot\ x+1)^2=0`
`⇔ cot\ x=-\frac{\sqrt{3}}{3}`
`⇔ x=-\frac{\pi}{3}+k\pi\ (k \in \mathbb{Z})`
