a.

\(1-sin^2x+1-2sin^2x+sinx+2=0\)

\(\Leftrightarrow-3sin^2x+sinx+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{4}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)

b. ĐKXĐ; ...

\(5tanx-\frac{2}{tanx}-3=0\)

\(\Leftrightarrow5tan^2x-3tanx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{2}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{2}{5}\right)+k\pi\end{matrix}\right.\)

e.

Ko rõ vế phải

f.

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow1-2sin^22x=0\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)

giúp mk với

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

=>\(1-\dfrac{3}{2}\cdot sin^22x=1-\dfrac{3}{4}sin^22x\)

=>\(sin^22x=0\)

=>2x=kpi

=>x=kpi/2

đây là câu a
mk cảm thấy cứ hơi sai sai . bạn xem lại hộ mk nhé

c/

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\)

\(\Leftrightarrow4sinx.cos2x+2cos2x=2\)

\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\)

\(\Leftrightarrow\left(1-2sin^2x\right)\left(2sinx+1\right)=1\)

\(\Leftrightarrow4sin^3x+2sin^2x-2sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+sinx-1\right)=0\)

\(\Leftrightarrow2sin^2x+sinx-1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b/

Nhận thấy \(sinx=0\) không phải nghiệm, pt tương đương:

\(sinx.cosx.cos2x.cos4x.cos8x=\frac{1}{16}sinx\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=sinx\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=sinx\)

\(\Leftrightarrow2sin8x.cos8x=sinx\)

\(\Leftrightarrow sin16x=sinx\)

\(\Rightarrow\left[{}\begin{matrix}16x=x+k2\pi\\16x=\pi-x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{15}\\x=\frac{\pi}{17}+\frac{k2\pi}{17}\end{matrix}\right.\)

\(cos^2x-\sqrt{3}sin2\text{x}=1+sin^2x\\ \Leftrightarrow cos2x-\sqrt{3}sin2\text{x}=1\\ \Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2\text{x}=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2\text{x}=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2\text{x}=\frac{1}{2}\\ \Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=\frac{\pi}{3}+a2\pi\\2x+\frac{\pi}{3}=-\frac{\pi}{3}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=a\pi\\x=-\frac{\pi}{3}+b\pi\end{matrix}\right.\)

câu 1:xét sinx=o

xét sinx khác 0

chia phương trình cho cos³x

ta được 1 phương trình mới:

4+3tanx-\(\frac{1}{sin^2x}\)-tan³x=0

<=>4+3tanx-(1+cot²x)-tan³x=0

<=>4+3tanx-1-\(\frac{1}{tan^2x}\)-tan³x=o

nhân cho tan²x ta được 1 phương trình bậc 5 với tanx