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a: \(\sin3x+cos2x=1+2\cdot\sin x\cdot cos2x\)
=>sin3x+cos2x=1+sin(x+2x)+sin(x-2x)
=>sin3x+cos2x=1+sin3x-sin x
=>cos2x-1+sin x=0
=>\(1-2\cdot\sin^2x-1+\sin x=0\)
=>\(-2\cdot\sin^2x+\sin x=0\)
=>sin x(2sin x-1)=0
TH1: sin x=0
=>\(x=k\pi\)
TH2: 2sin x-1=0
=>\(\sin x=\frac12\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{6}+k2\pi\\ x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\)
b: \(\sin^3x+cos^3x=2\cdot\left(\sin^5x+cos^5x\right)\)
=>\(\sin^3x-2\cdot\sin^5x+cos^3x-2\cdot cos^5x=0\)
=>\(\sin^3x\left(1-2\cdot\sin^2x\right)+cos^3x\left(1-2\cdot cos^2x\right)=0\)
=>\(\sin^3x\cdot cos2x-cos^3x\cdot cos2x=0\)
=>\(cos2x\left(\sin^3x-cos^3x\right)=0\)
TH1: cos2x=0
=>\(2x=\frac{\pi}{2}+k\pi\)
=>\(x=\frac{\pi}{4}+\frac{k\pi}{2}\)
TH2: \(\sin^3x-cos^3x=0\)
=>\(\sin^3x=cos^3x\)
=>sin x=cosx
=>\(\sin x-cosx=0\)
=>\(\sqrt2\cdot\sin\left(x-\frac{\pi}{4}\right)=0\)
=>\(\sin\left(x-\frac{\pi}{4}\right)=0\)
=>\(x-\frac{\pi}{4}=k\pi\)
=>\(x=\frac{\pi}{4}+k\pi\)
f: ĐKXĐ: \(\begin{cases}\sin x<>0\\ cosx<>0\end{cases}\Rightarrow\begin{cases}x<>k\pi\\ x<>\frac{\pi}{2}+k\pi\end{cases}\Rightarrow x<>\frac{k\pi}{2}\)
\(\frac{\tan x-\sin x}{\sin^3x}=\frac{1}{cosx}\)
=>\(\frac{\frac{\sin x}{cosx}-\sin x}{\sin^3x}=\frac{1}{cosx}\)
=>\(\frac{\frac{1}{cosx}-1}{\sin^2x}=\frac{1}{cosx}\)
=>\(\sin^2x=cosx\cdot\left(\frac{1}{cosx}-1\right)=1-cosx\)
=>\(1-cos^2x=1-cosx\)
=>\(cos^2x-cosx=0\)
=>cosx(cosx-1)=0
TH1: cosx=0
=>\(x=\frac{\pi}{2}+k\pi\) (loại)
TH2: cosx-1=0
=>cosx=1
=>\(x=k2\pi\)
=>sin x=0
=>Loại
1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)