Biến đổi pt trên như sau:

sinx.cosx/4 + cosx.sinx/4 - 3(sin²x + cos²x)  + cosx = 0

sin(x + x/4) + cosx = 3

sin5x/4 + cosx = 3

Vì sin5x/4 \(\le\) 1 và cosx \(\le\) 1. Do đó sin5x/4 + cosx \(\le\) 2. Vì vậy pt trên vô nghiệm. 

\(pt\Leftrightarrow\cos\frac{x}{4}\sin x+\cos x+\sin\frac{x}{4}\cos x=3\left(\sin^2x+\cos^2x\right)=3\)

Mà \(\sin\alpha;\text{ }\cos\alpha\le1\forall\alpha\)

\(\Rightarrow\cos\frac{x}{4}.\sin x\le1.1;\text{ }\sin\frac{x}{4}.\cos x\le1.1;\text{ }\cos x\le1\forall x\)

\(\Rightarrow\cos\frac{x}{4}.\sin x+\sin\frac{x}{4}.\cos x+\cos x\le3\text{ }\forall x\)

Dấu "=" xảy ra khi \(\cos x=1;\text{ }\cos\frac{x}{4}.\sin x=1;\text{ }\cos x.\sin\frac{x}{4}=1\)

\(\Leftrightarrow\cos x=1;\text{ }\sin\frac{x}{4}=1;\text{ }\cos\frac{x}{4}.\sin x=1\)

Pt trên vô nghiệm do \(\cos x=1\text{ thì }\sin x=0\Rightarrow\cos\frac{x}{4}.\sin x=0\)

Vậy phương trình đã cho vô nghiệm.

a/ Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow2tan^3x=\frac{1}{cos^2x}\)

\(\Leftrightarrow2tan^3x=1+tan^2x\)

\(\Leftrightarrow2tan^3x-tan^2x-1=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(2tan^2x+tanx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\2tan^2x+tanx+1=0\left(vn\right)\end{matrix}\right.\)

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

b/

\(\Leftrightarrow sin^3x-sinx+cos^3x+cosx=0\)

\(\Leftrightarrow-sinx\left(1-sin^2x\right)+cos^3x+cosx=0\)

\(\Leftrightarrow-sinx.cos^2x+cos^3x+cosx=0\)

\(\Leftrightarrow cosx\left(cos^2x-sinx.cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\cos^2x-sinx.cosx+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos2x-\frac{1}{2}sin2x+1=0\)

\(\Leftrightarrow cos2x-sin2x=-3\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=\frac{3}{\sqrt{2}}>1\left(vn\right)\)

c/

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(4+2tan^3x-3tanx.\frac{1}{cos^2x}=0\)

\(\Leftrightarrow2tan^3x-3tanx\left(1+tan^2x\right)+4=0\)

\(\Leftrightarrow-tan^3x-3tanx+4=0\)

\(\Leftrightarrow\left(1-tanx\right)\left(tan^2x+tanx+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x+tanx+4=0\left(vn\right)\end{matrix}\right.\)

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

d/

\(\Leftrightarrow2cos^3x+2sinx-6sin^2x.cosx=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(2+2tanx.\frac{1}{cos^2x}-6tan^2x=0\)

\(\Leftrightarrow1+tanx\left(1+tan^2x\right)-3tan^2x=0\)

\(\Leftrightarrow tan^3x-3tan^2x+tanx+1=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x-2tanx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x-2tanx-1=0\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)

a/

\(\Leftrightarrow cos^3x-sin^3x=cosx+sinx\)

- Với \(cosx=0\Rightarrow sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\) là 1 nghiệm

- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow1-tan^3x=\frac{1}{cos^2x}+tanx.\frac{1}{cos^2x}\)

\(\Leftrightarrow1-tan^3x=1+tan^2x+tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+tan^2x+tanx=0\)

\(\Leftrightarrow tanx\left(2tan^2x+tanx+1\right)=0\)

\(\Leftrightarrow tanx=0\Rightarrow x=k\pi\)

b/

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}=1+2sinx\)

\(\Leftrightarrow\frac{cosx-sinx}{cosx+sinx}=1+2sinx\)

\(\Leftrightarrow cosx-sinx=\left(1+2sinx\right)\left(cosx+sinx\right)\)

\(\Leftrightarrow sinx+sinx.cosx+sin^2x=0\)

\(\Leftrightarrow sinx\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx+cosx=-1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\left(l\right)\\x=\pi+k2\pi\end{matrix}\right.\)

c/

ĐKXĐ: ...

Chia 2 vế cho \(cos^2x\) ta được:

\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)

\(\Leftrightarrow tan^3x+tan^2x=3tanx-3tan^2x+3+3tan^2x\)

\(\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

a.

a/

\(\Leftrightarrow cos2x=sin3x\)

\(\Leftrightarrow cos2x=cos\left(\frac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-3x+k2\pi\\2x=3x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k2\pi}{5}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

b/

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{2}\\sinx=1-\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow x=...\)

c/

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1+cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow1-3sin^2x.cos^2x=1+sin2x\)

\(\Leftrightarrow-\frac{3}{4}sin^22x=sin2x\)

\(\Leftrightarrow3sin^22x+4sin2x=0\)

\(\Leftrightarrow sin2x\left(3sin2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=-\frac{4}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\frac{k\pi}{2}\)

1.

Đề là \(x\in\left(0;\frac{\pi}{4}\right)\) hay \(x\in\left[0;\frac{\pi}{4}\right]\) ?

2.

\(sin3x-4sinx.cos2x=0\)

\(\Leftrightarrow sin3x-\left(2sin3x-2sinx\right)=0\)

\(\Leftrightarrow2sinx-sin3x=0\)

\(\Leftrightarrow2sinx-3sinx+4sin^3x=0\)

\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)

\(\Leftrightarrow sinx\left(1-2cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

3.

\(sin^2x.cosx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

4.

\(\sqrt{3}sin2x+1-cos2x=3\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{3}+k\pi\)

5.

Ko có 4 đáp án thì làm sao biết, có vô số pt tương đương với pt này :)

6.

\(sinx+cosx-2sinx.cosx+1=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t+1-t^2+1=0\)

\(\Leftrightarrow-t^2+t+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2sinx.cosx=t^2-1=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)