$\Rightarrow -\sin x \cos2x – \cos2x \cos x + 2\cos2x = 0$
$\Rightarrow \cos2x (-\sin x – \cos x + 2) = 0$
`=>`\(\left[ \begin{array}{l}\text{cos }2x=0\\-\sin x – \cos x + 2=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}4+\dfrac{k\pi}2\\\sin x + \cos x = 2\end{array} \right.\)
Đáp án:
Giải thích các bước giải:
`tanx -sin2x- cos2x+ 2(2cosx-\frac{1}{cos x} )= 0`
ĐK: `x \ne \frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})`
`⇔ \frac{sin x}{cos x}-sin 2x- cos 2x +4 cos x-\frac{2}{cos x}=0`
`⇔ sin x-2sin x cos^2 x-cos 2x cos x-2(cos^2x-1)=0`
`⇔ sin x(1-2cos^2x)-cos 2x cos x+2cos2x=0`
`⇔ -sin xcos2x-cos 2x cos x+2cos2x=0`
`⇔ cos 2x(sin x+cos x-2)=0`
`⇔` \(\left[ \begin{array}{l}cos 2x=0\\sin x+cos x=2\ (vô\ nghiệm)\end{array} \right.\)`
`⇒ x=\frac{\pi}{4}+k\frac{\pi}{2}\ (TM)\ (k \in \mathbb{Z})`
Đáp án:
`x=\frac{\pi}4+\frac{k\pi}2` `(kinZZ)`
Giải thích các bước giải:
`tanx sin2x- cos2x+ 2(2cosx-\frac{1}{cos x} )= 0`
Nhân `2` vế với `cosx`
$\Rightarrow\sin x – \sin2x \cos x – \cos2x \cos x + 2(2\cos^2x – 1) = 0$
$\Rightarrow \sin x – 2\sin x \cos x. \cos x – \cos2x \cos x + 2 \cos2x = 0$
$\Rightarrow \sin x(1 – 2 \cos^2x) – \cos2x \cos x + 2\cos2x = 0$
$\Rightarrow -\sin x \cos2x – \cos2x \cos x + 2\cos2x = 0$
$\Rightarrow \cos2x (-\sin x – \cos x + 2) = 0$
`=>`\(\left[ \begin{array}{l}\text{cos }2x=0\\-\sin x – \cos x + 2=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}4+\dfrac{k\pi}2\\\sin x + \cos x = 2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}4+\dfrac{k\pi}2\\\sqrt{2} \sin(x + \dfrac{\pi}{4}) = 2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}4+\dfrac{k\pi}2\\\sin(x + \dfrac{\pi}{4}) = \sqrt{2}\text{ (loại do }\sqrt2>1)\end{array} \right.\)
`=>x=\frac{\pi}4+\frac{k\pi}2` `(kinZZ)`
Vậy `x=\frac{\pi}4+\frac{k\pi}2` `(kinZZ).`