Câu hỏi của Julian Edward

Question

giải các pt

a) $5\left(1+cosx\right)=2+sin^4x-cos^4x$

b) $\sqrt{3}tanx+cotx-\sqrt{3}-1=0$

c) $6sin^2x+2sin^22x=5$

d) $cos^22x+cos^2\left(x-\frac{\pi}{4}\right)-1=0$

e) \(\left(1+tan^2x\ri...

Nguyễn Việt Lâm · Accepted Answer

a/

$\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)$

$\Leftrightarrow3+5cosx=sin^2x-cos^2x$

$\Leftrightarrow3+5cosx=\left(1-cos^2x\right)-cos^2x$

$\Leftrightarrow2cos^2x+5cosx+2=0$

$\Rightarrow\left[{}\begin{matrix}cosx=-2\left(l\right)\cosx=-\frac{1}{2}\end{matrix}\right.$

$\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi$

b/ ĐKXĐ: ...

$\Leftrightarrow\sqrt{3}tanx+\frac{1}{tanx}-\sqrt{3}-1=0$

$\Leftrightarrow\sqrt{3}tan^2x-\left(\sqrt{3}+1\right)tanx+1=0$

$a+b+c=\sqrt{3}-\left(\sqrt{3}+1\right)+1=0$

$\Rightarrow\left[{}\begin{matrix}tanx=1\tanx=\frac{1}{\sqrt{3}}\end{matrix}\right.$ $\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\x=\frac{\pi}{6}+k\pi\end{matrix}\right.$Câu trả lời: a/

$\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)$

$\Leftrightarrow3+5cosx=sin^2x-cos^2x$

$\Leftrightarrow3+5cosx=\left(1-cos^2x\right)-cos^2x$

$\Leftrightarrow2cos^2x+5cosx+2=0$

$\Rightarrow\left[{}\begin{matrix}cosx=-2\left(l\right)\cosx=-\frac{1}{2}\end{matrix}\right.$

$\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi$

b/ ĐKXĐ: ...

$\Leftrightarrow\sqrt{3}tanx+\frac{1}{tanx}-\sqrt{3}-1=0$

$\Leftrightarrow\sqrt{3}tan^2x-\left(\sqrt{3}+1\right)tanx+1=0$

$a+b+c=\sqrt{3}-\left(\sqrt{3}+1\right)+1=0$

$\Rightarrow\left[{}\begin{matrix}tanx=1\tanx=\frac{1}{\sqrt{3}}\end{matrix}\right.$ $\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\x=\frac{\pi}{6}+k\pi\end{matrix}\right.$

Nguyễn Việt Lâm · Answer

c/

$\Leftrightarrow6\left(\frac{1-cos2x}{2}\right)+2\left(1-cos^22x\right)=5$

$\Leftrightarrow-2cos^22x-3cos2x=0$

$\Leftrightarrow cos2x\left(2cos2x+3\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}cos2x=0\cos2x=-\frac{3}{2}\left(l\right)\end{matrix}\right.$

$\Rightarrow2x=\frac{\pi}{2}+k\pi$

$\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}$Câu trả lời: c/

$\Leftrightarrow6\left(\frac{1-cos2x}{2}\right)+2\left(1-cos^22x\right)=5$

$\Leftrightarrow-2cos^22x-3cos2x=0$

$\Leftrightarrow cos2x\left(2cos2x+3\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}cos2x=0\cos2x=-\frac{3}{2}\left(l\right)\end{matrix}\right.$

$\Rightarrow2x=\frac{\pi}{2}+k\pi$

$\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}$

Nguyễn Việt Lâm · Answer

d/

$\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0$

$\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0$

$\Leftrightarrow-2sin^22x+sin2x+1=0$

$\Rightarrow\left[{}\begin{matrix}sin2x=1\sin2x=-\frac{1}{2}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\2x=-\frac{\pi}{6}+k2\pi\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\x=-\frac{\pi}{12}+k\pi\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.$Câu trả lời: d/

$\Leftrightarrow cos^22x+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{\pi}{2}\right)-1=0$

$\Leftrightarrow1-sin^22x+\frac{1}{2}sin2x-\frac{1}{2}=0$

$\Leftrightarrow-2sin^22x+sin2x+1=0$

$\Rightarrow\left[{}\begin{matrix}sin2x=1\sin2x=-\frac{1}{2}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\2x=-\frac{\pi}{6}+k2\pi\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\x=-\frac{\pi}{12}+k\pi\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.$

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