mn giúp em giải hai phương trình sau với ạ :,hứa vote 5 sao+cám ơn + câu tlhn
1) Sin^2 2x – 2cos^2 x +
$\frac{3}{4}$ = 0
2) 5Sinx – 2 = 3(1-sinx).tan^2 x
mn giúp em giải hai phương trình sau với ạ :,hứa vote 5 sao+cám ơn + câu tlhn
1) Sin^2 2x – 2cos^2 x +
$\frac{3}{4}$ = 0
2) 5Sinx – 2 = 3(1-sinx).tan^2 x
Đáp án:
1) $x = \pm \dfrac{\pi}{6} + k\pi\quad (k\in\Bbb Z)$
2) $\left[\begin{array}{l}x = \dfrac{\pi}{6}+ k\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.(k\in\Bbb Z)$
Giải thích các bước giải:
1) $\sin^22x – 2\cos^2x + \dfrac{3}{4}=0$
$\Leftrightarrow 1 – \cos^22x – 2.\dfrac{1 + \cos2x}{2} + \dfrac{3}{4}=0$
$\Leftrightarrow \cos^22x + \cos2x – \dfrac{3}{4} = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos2x = \dfrac{1}{2}\\\cos2x = -\dfrac{3}{2}\quad (loại)\end{array}\right.$
$\Leftrightarrow 2x = \pm \dfrac{\pi}{3} + k2\pi$
$\Leftrightarrow x = \pm \dfrac{\pi}{6} + k\pi\quad (k\in\Bbb Z)$
2) $5\sin x – 2 = 3(1 – \sin x)\tan^2x \qquad (*)$
$ĐKXĐ:\, x \ne \dfrac{\pi}{2} + n\pi$
$(*)\Leftrightarrow 5\sin x – 2 = 3(1 – \sin x)\dfrac{\sin^2x}{\cos^2x}$
$\Leftrightarrow 5\sin x – 2 = 3(1 – \sin x)\dfrac{\sin^2x}{1 -\sin^2x}$
$\Leftrightarrow 5\sin x – 2 = 3(1 – \sin x)\dfrac{\sin^2x}{(1 -\sin x)(1 + \sin x)}$
$\Leftrightarrow 5\sin x – 2 = 3\dfrac{\sin^2x}{1 +\sin x}$
$\Leftrightarrow (5\sin x – 2)(1 + \sin x) = 3\sin^2x$
$\Leftrightarrow 2\sin^2x + 3\sin x – 2 = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = \dfrac{1}{2}\\\sin x = -2\quad (loại)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6}+ k\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.(k\in\Bbb Z)$
`1) sin^2 2x – 2cos^2 x + 3/4 = 0`
`<=> 1 – cos^2 2x – 2.(1 + cos 2x)/2 + 3/4 = 0`
`<=> 1 – cos^2 2x – 1 – cos 2x + 3/4 = 0`
`<=> -cos^2 2x – cos 2x + 3/4 = 0`
`<=> -4cos^2 2x – 4cos 2x + 3 = 0`
`<=>` \(\left[ \begin{array}{l}cos 2x = \dfrac{1}{2}\\cos 2x = -\dfrac{3}{2} (l)\end{array} \right.\)
`<=> x = ±(π)/6 + kπ` `(k ∈ ZZ)`
`2) 5sin x – 2 = 3(1 – sin x).tan^2 x` `(D = RR \\ {π/2 + kπ | k ∈ ZZ)`
`<=> 5sin x – 2 = 3(1 – sin x).(sin^2 x)/((1 – sin x)(1 + sin x))`
`=> (5sin x – 2)(1 + sin x) = 3sin^2 x`
`<=> 5sin x + 5sin^2 x – 2 – 2sin x – 3sin^2 x = 0`
`<=> 2sin^2 x + 3sin^2 x – 2 = 0`
`<=>` \(\left[ \begin{array}{l}sin x = \dfrac{1}{2}\\sin x = -2 (l)\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)`