Question

Giải các Phương trình sau

a) \(sin^4\frac{x}{2}+cos^4\frac{x}{2}=\frac{1}{2}\)

b) \(sin^6x+cos^6x=\frac{7}{16}\)

c) \(sin^6x+cos^6x=cos^22x+\frac{1}{4}\)

d) \(tanx=1-cos2x\)

e) \(tan(2x+\frac\pi3).tan(\frac\pi3-x)=1\)

f) \(tan(x-15^o).cot(x+15^o)=\frac{1}{3}\)

Answer 1

a.

\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

\(\Leftrightarrow cos^2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

Answer 2

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

Answer 3

c.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow3-3sin^22x=4cos^22x\)

\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)

\(\Leftrightarrow3=3+cos^22x\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Answer 4

d.

ĐKXD: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(\dfrac{sinx}{cosx}=1-\left(1-2sin^2x\right)\)

\(\Leftrightarrow2sin^2x-\dfrac{sinx}{cosx}=0\)

\(\Leftrightarrow sinx\left(2sinx-\dfrac{1}{cosx}\right)=0\)

\(\Leftrightarrow\dfrac{sinx\left(2sinx.cosx-1\right)}{cosx}=0\)

\(\Leftrightarrow\dfrac{sinx\left(sin2x-1\right)}{cosx}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

Answer 5

e.

ĐKXĐ: \(\left\{{}\begin{matrix}2x+\dfrac{\pi}{3}\ne\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{3}\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\x\ne\dfrac{5\pi}{6}+k\pi\end{matrix}\right.\)

\(tan\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{tan\left(\dfrac{\pi}{3}-x\right)}\)

\(\Rightarrow tan\left(2x+\dfrac{\pi}{3}\right)=cot\left(\dfrac{\pi}{3}-x\right)\)

\(\Leftrightarrow tan\left(2x+\dfrac{\pi}{3}\right)=tan\left(\dfrac{\pi}{6}+x\right)\)

\(\Leftrightarrow2x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+x+k\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi\) (loại)

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Answer 6

f.

ĐKXĐ: \(\left\{{}\begin{matrix}x-15\ne90+k180\\x+15\ne90+k180\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne105+k180\\x\ne75+k180\end{matrix}\right.\)

\(\dfrac{sin\left(x-15^0\right)}{cos\left(x-15^0\right)}.\dfrac{cos\left(x+15^0\right)}{sin\left(x+15^0\right)}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{sin2x+sin\left(-30^0\right)}{sin2x+sin\left(30^0\right)}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{sin2x-\dfrac{1}{2}}{sin2x+\dfrac{1}{2}}=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{2sin2x-1}{2sin2x+1}=\dfrac{1}{3}\)

\(\Rightarrow6sin2x-3=2sin2x+1\)

\(\Leftrightarrow sin2x=1\)

\(\Rightarrow x=45^0+k180^0\)