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2sin^x + sin2x – 2cos^x = 1 Giúp tui với pleaseeeeee
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Đáp án: $\text{\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+ k\pi \\x=arctan(-3)+ k\pi\end{array} \right.\) (k∈Z)}$
Giải thích các bước giải:
$\text{2sin$^{2}$ + sin2x – 2cos$^{2}$x= 1 }$
⇔ $\text{2sin$^{2}$ + 2sinx.coss – 2cos$^{2}$x= 1 }$
$\text{* cos x=0 ⇒x=$\dfrac{\pi}{2}$ + k2$\pi$ }$
$\text{pt⇒ 2=1 (sai)}$
vậy $\text{x=$\dfrac{\pi}{2}$ + k2$\pi$ }$ không là nghiệm của pt.
$\text{* cos x $\neq$ 0 ⇒x$\neq$$\dfrac{\pi}{2}$ + k2$\pi$, k∈z, chia hai vế cho cos$^{2}$x ta được }$
$\text{2tan$^{2}$x + 2tanx – 2= 1.(1+tan$^{2}$x)}$
⇔ $\text{2tan$^{2}$x + 2tanx – 2 – 1 – tan$^{2}$x=0}$
⇔$\text{tan$^{2}$x + 2tanx – 3 =0}$
⇔$\text{\(\left[ \begin{array}{l}tan x=1\\tan x=-3\end{array} \right.\) }$
⇒$\text{\(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+ k\pi \\x=arctan(-3)+ k\pi\end{array} \right.\) (k∈Z)}$